PAS MATEMATIKA BLOG

Contoh Soal 2 Turunan Fungsi Aljabar

$\begin{array}{ll}\\ 6.&\textrm{Jika}\: \: W=\sin 2t\: ,\: \textrm{maka}\: \: \displaystyle \frac{dW}{dt}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos 2t&\textrm{d}.\quad 2t\cos 2t+\sin 2t\\ \textrm{b}.\quad \color{red}2\cos 2t\quad &\textrm{e}.\quad \sin 2t-t\cos 2t\\ \textrm{c}.\quad \sin 2t+t\cos 2t\quad \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &W=\sin 2t\\ &W=\sin u\, \quad \textrm{dengan}\: \: u=2t\\ &\displaystyle \frac{dW}{dt}=\displaystyle \frac{dW}{du}.\frac{du}{dt}\\ &\qquad=\cos u.2\\ &\qquad=2\cos u\\ &\qquad=2\cos 2t \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{2x+4}{1+\sqrt{x}}\: ,\: \textrm{maka}\: \: \displaystyle {f}\, '(4) =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{4} &&\textrm{d}.\quad 1 \\ \textrm{b}.\quad \displaystyle \frac{3}{7} \quad &\textrm{c}.\quad \displaystyle \frac{3}{5} \quad &\textrm{e}.\quad 4 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{2x+4}{1+\sqrt{x}}\\ &=\displaystyle \frac{U}{V}\\ {f}\, '(x)&=\displaystyle \frac{{U}\, '.V-U.{V}\, '}{V^{2}}\\ &=\displaystyle \frac{(2)\left ( 1+\sqrt{x} \right )-(2x+4).\left ( 1.\left ( 1+\sqrt{x} \right )^{0} .\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )}{\left ( 1+\sqrt{x} \right )^{2}} \\&\textrm{ingat}\: \: \sqrt{x}=x^{^{\frac{1}{2}}} \\ {f}\, '(4) &=\displaystyle \frac{2\left ( 1+\sqrt{4} \right )-(2.4+4).\frac{1}{2}. 4^{^{-\frac{1}{2}}} }{\left ( 1+\sqrt{4} \right )^{2}}\\ &=\displaystyle \frac{2(1+2)-(12).\frac{1}{2}. \frac{1}{2}}{(1+2)^{2}}\\ &\textrm{ingat juga}\: \: 4^{^{-\frac{1}{2}}}=\left ( 2^{2} \right )^{^{-\frac{1}{2}}}=2^{^{-1}}=\displaystyle \frac{1}{2^{1}}=\frac{1}{2}\\ &=\displaystyle \frac{6-3}{9}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x} \: ,\: \textrm{maka}\: \: {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2} &&\textrm{d}.\quad \color{red}2 \\ \textrm{b}.\quad \displaystyle -\frac{1}{2} \quad &\textrm{c}.\quad \displaystyle \frac{3}{4} \quad &\textrm{e}.\quad -2 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{1+\sin x}{\cos x}\\ {f}\, '(x)&=\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x +\sin ^{2}x}{\cos ^{2}x}\\ &\qquad \textrm{ingat bahwa}\: \: \sin ^{2}x+\cos ^{2}x=1\\ &=\displaystyle \frac{1+\sin x }{\cos ^{2}x}\\ {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )&=\displaystyle \frac{1+\sin \left ( \displaystyle \frac{1}{6}\pi \right )}{\cos ^{2}\left ( \displaystyle \frac{1}{6}\pi \right )}\\&=\displaystyle \frac{1+\displaystyle \frac{1}{2} }{\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\displaystyle \frac{3}{2}}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{3}{2}\times \frac{4}{3}\\ &=2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: \: f(x)=\displaystyle 3x^{2}-2ax+7\\ &\textrm{dan}\: \: {f}\, '(1)=0 \: ,\: \textrm{maka}\: \: {f}\, '(2)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1 &&\textrm{d}.\quad \color{red}6 \\ \textrm{b}.\quad \displaystyle 2 \quad &\textrm{c}.\quad \displaystyle 4 \quad &\textrm{e}.\quad 8 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle 3x^{2}-2ax+7\\ {f}\, '(x)&=6x-2a\\ {f}\, '(1)&=0\\ 6(1)-2a&=0\\ 6&=2a\\ 3&=a\\ \textrm{sehingga}\, &\: \\ {f}\, '(x)&=6x-6\\ \textrm{maka}\, ,\: \quad &\\ {f}\, '(2)&=6.2-6\\ &=12-6\\ &=6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: \: f(x)=\displaystyle (6x-3)^{3}(2x-1)\\ &\textrm{maka}\: \: {f}\, '(1)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 18 &&\textrm{d}.\quad 162 \\\\ \textrm{b}.\quad \displaystyle 24 \quad &\textrm{c}.\quad \displaystyle 54 \quad &\textrm{e}.\quad \color{red}216 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle (6x-3)^{3}(2x-1)\\ &=(3.(2x-1))^{3}(2x-1)^{1}\\ &=3^{3}.(2x-1)^{3+1}\\ &=27(2x-1)^{4}\\ {f}\, '(x)&=4.27(2x-1)^{4-1}.2\\ &=216.(2x-1)^{3}\\ {f}\, '(1)&=216.(2.1-1)^{3}\\ &=216.1\\ &=216 \end{aligned} \end{array}$.

Contoh Soal 1 Turunan Fungsi Aljabar

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: {f}\, '(2)=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-8}{x-2},\\ & \textrm{maka fungsi}\: \: f(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 12&&\textrm{d}.\quad \color{red}x^{3}\\ \textrm{b}.\quad 2\quad&\textrm{c}.\quad x\quad&\textrm{e}.\quad x-8 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned} &\textrm{Turunan fungsi f di}\: \: x=c\: \: \textrm{adalah}\\ &f\, '(c)=\underset{x\rightarrow c}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(c)}{x-c},\: \: \textrm{maka}\\ &\textrm{turunan fungsi f di}\: \: x=2\: \: \textrm{adalah}\\ &f\, '(2)=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-8}{x-2}\\ &\textrm{sehingga akan didapa}\textrm{tkan fungsi}\: \: f\: \: \textrm{nya }\\ &\textrm{yaitu}\: \: f(x)=\color{red}x^{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: a\neq 0\, ,\: \textrm{maka nilai dari}\\&\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3a\displaystyle \sqrt[3]{a}&&\textrm{d}.\quad \displaystyle \frac{1}{2a}\sqrt[3]{a} \\ \textrm{b}.\quad 2a\displaystyle \sqrt[3]{a} \quad&\textrm{c}.\quad 0\quad&\textrm{e}.\quad \color{red}\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{array}\\\\ &\textbf{Jawab}:\\ &\color{blue}\textrm{Alternatif 1}\\ &\begin{aligned}{f}\, '(a)&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{\left ( \sqrt[3]{x}-\sqrt[3]{a} \right )\left ( \sqrt[3]{x^{2}}+\sqrt[3]{xa}+\sqrt[3]{a^{2}} \right )}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt[3]{x^{2}}+\sqrt[3]{xa}+\sqrt[3]{a^{2}} \right )}\\ &=\displaystyle \frac{1}{\left ( \sqrt[3]{a^{2}}+\sqrt[3]{a.a}+\sqrt[3]{a^{2}} \right )}\\ &=\displaystyle \frac{1}{3\sqrt[3]{a^{2}}}\\ &=\displaystyle \frac{1}{3\sqrt[3]{a^{2}}}\times \displaystyle \frac{\sqrt[3]{a}}{\sqrt[3]{a}}\\ &=\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{aligned}\\ &\color{blue}\textrm{Alternatif 2 (dengan aturan L'Hopital)}\\&\begin{aligned}\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)}{g(x)}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{{f}\, '(x) }{{g}\, '(x) }\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{x^{^{\frac{1}{3}}}-a^{^{\frac{1}{3}}} }{x-a}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{3}x^{^{\frac{1}{3}-1}}\: -0}{1-0}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{3}x^{^{-\frac{2}{3}}}}{1}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{3\sqrt[3]{x^{2}}}\times \frac{\sqrt[3]{x}}{\sqrt[3]{x}}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{3x}\sqrt[3]{x}\\ &=\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{aligned} \end{array}$ .

$\begin{array}{ll}\\ 3.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x}}\: \: \textrm{maka nilai dari}\\ & -2{f}\, '(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{x\sqrt{x}} &&\textrm{d}.\quad \displaystyle \color{red}-\frac{1}{2x\sqrt{x}} \\ \textrm{b}.\quad \displaystyle x\sqrt{x} \quad&\textrm{c}.\quad \displaystyle -\frac{1}{2\sqrt{x}} \quad&\textrm{e}.\quad \displaystyle -2x\sqrt{x} \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}\textrm{Dike}&\textrm{tahui}\\ f(x)&=\displaystyle \frac{1}{\sqrt{x}}=\displaystyle \frac{1}{x^{^{\frac{1}{2}}}}=x^{^{-\frac{1}{2}}} \end{aligned}\\ &\begin{array}{|c|c|}\hline y=ax^{n}\rightarrow {y}\, '=nax^{n-1}&\begin{aligned}&\\ y&=\displaystyle \frac{U}{V}\rightarrow {y}\, '=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}}\\ & \end{aligned}\\\hline \begin{aligned} {f}\, '(x)&=-\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{2}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{2x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{2x^{1}.x^{^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{1}{2x\sqrt{x}} \end{aligned}&\begin{aligned} {f}\, '(x)&=\displaystyle \frac{0.\sqrt{x}-1.\frac{1}{2}x^{^{\frac{1}{2}-1}}}{\left ( \sqrt{x}\right )^{2}}\\ &=\displaystyle \frac{-\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}}{x}\\ &=-\displaystyle \frac{1}{2xx^{^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{1}{2x\sqrt{x}} \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Turunan pertama dari}\: \: y=\displaystyle \sqrt[n]{x}\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{n}x^{^{\frac{1}{n}}}&&\textrm{d}.\quad (n-1)\sqrt[n-1]{x}\\ \textrm{b}.\quad \displaystyle \color{red}\frac{1}{n}x^{^{\frac{1-n}{n}}}\quad &\textrm{c}.\quad \displaystyle \frac{1}{n-1}x^{^{n-1}}\quad &\textrm{e}.\quad \sqrt[n-1]{x} \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}y&=\sqrt[n]{x}=\displaystyle x^{^{\frac{1}{n}}},\: \: \textrm{maka}\\ {y}\, '&=\displaystyle \frac{1}{n}x^{^{\frac{1}{n}-1}}=\displaystyle \frac{1}{n}x^{^{\frac{1-n}{n}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Turunan ke}-n\: \: \textrm{dari}\: \: \: y=\displaystyle \frac{1}{x} \: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle n!\: .\: x^{-(n+1)} & \\ \textrm{b}.\quad \displaystyle (n+1)!\: .\: x^{-(n+1)}& \\ \textrm{c}.\quad \color{red}\displaystyle (-1)^{n}\: n!\: .\: x^{-(n+1)}\\ \textrm{d}.\quad (-1)^{n+1}\: n!\: .\: x^{-(n+1)}\\ \textrm{e}.\quad (-1)^{n+1}\: (n+1)!\: .\: x^{-(n+1)} \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|l|}\hline \textrm{Fungsi}&y=\displaystyle \frac{1}{x}&\quad y=x^{-1}\\\hline y'=\displaystyle \frac{dy}{dx}&-x^{-2}&(-1)^{1}.1!.x^{-(1+1)}\\\hline y''=\displaystyle \frac{d^{2}y}{dx^{2}}&2x^{-3}&(-1)^{2}.2!.x^{-(2+1)}\\\hline y'''=\displaystyle \frac{d^{3}y}{dx^{3}}&-6x^{-4}&(-1)^{3}.3!.x^{-(3+1)}\\\hline y^{IV}=\displaystyle \frac{d^{4}y}{dx^{4}}&24x^{-5}&(-1)^{4}.4!.x^{-(4+1)}\\\hline y^{V}=\displaystyle \frac{d^{5}y}{dx^{5}}&\cdots &\quad \cdots \\\hline \vdots &\vdots &\quad \vdots \\\hline y^{n}=\displaystyle \frac{d^{n}y}{dx^{n}}&\cdots &\color{red}(-1)^{n}.n!.x^{-(n+1)}\\\hline \end{array} \end{array}$.

Menyelesaikan Masalah Berkaitan Keekstriman Fungsi dan Penggunaan Turunan Kedua Fungsi Aljabar.

Lanjutan Materi Penggunaan Turunan Fungsi (Silahkan Lihat materi sebelumnya di sini)

$\begin{array}{ll}\\ 5.&\textrm{Diketahui jumlahdua bilangan adalah 30}\\ &\textrm{Jika perkalian salah satu bilangan dengan}\\ &\textrm{kuadrat bilangan lainnya mencapai nilai}\\ &\textrm{maksimum, tentukanlah}:\\ &\textrm{a}.\quad \textrm{kedua bilangan tersebut}\\ &\textrm{b}.\quad \textrm{nilai maksimumnya}\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan salah satu bilangan adalah}\: \: x,\\ &\textrm{maka bilangan yang lainnya adalah}:\: \: 30-x\\ &\textrm{Dan misalkan pula perkalian ini dirumuskan}\\ &\textrm{dengan}\: \: p(x),\: \textrm{maka}\\ &\begin{aligned}\textrm{a}.\quad p(x)\: &=(30-x)x^{2}=30x^{2}-x^{3}\\ \textrm{syara}&\textrm{t ekstrim maksimum}\: \: p'(x)=0\\ 60x-&3x^{2}=0\Leftrightarrow 3x(20-x)=0\\ x_{1}=\: &0\: \: \textrm{atau}\: \: x_{2}=20\\ \textrm{Jadi},&\: \textrm{nilai agar maksimum}\: \: x=20\: \: \textrm{dan}\\ \textrm{nilai}&\: \: x\: \: \textrm{yang lain adalah}:30-20=10\\ \textrm{b}.\quad \textrm{Dan}\: &\textrm{nilai maksimumnya adalah}:\\ p(\color{red}20\color{black})&=(30-20).20^{2}=10.400=\color{red}4000\: \: (\textbf{mak})\\ \end{aligned} \end{array}$ .

$\begin{array}{ll}\\ 6.&\textrm{Sebuah peluru ditembakkan ke atas. Dan}\\ &\textrm{tinggi yang dapat dicapai peluru adalah}\\ &h\: \: \textrm{meter dalam waktu}\: \: t\: \: \textrm{detik yang dapat}\\ &\textrm{rumuskan dengan}\: \: h(t)=120t-5t^{2}\\ &\textrm{Tentukanlah}\\ &\textrm{a}.\quad t\: \: \textrm{agar}\: \: h\: \: \textrm{maksimum}\\ &\textrm{b}.\quad \textrm{tinggi}\: \: h\: \: \textrm{maksimum}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \textrm{syara}&\textrm{at ekstrim maksimum}\: \: h'(x)=0\\ 120-&10t=0\Leftrightarrow t=\color{red}12\\ \textrm{Jadi},&\: \textrm{tinggi maksimum dicapai saat}\: \: \color{red}t=12\\ \textrm{b}.\quad \textrm{Dan}\: &\textrm{tinggi maksimumnya adalah}:\\h(\color{red}12\color{black})&=120(12)-5(12)^{2}=\color{blue}720\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&(\textbf{OSK 2018})\\ &\textrm{Diketahui bilangan real}\: \: x\: \: \textrm{dan}\: \: y\\ &\textrm{yang memenuhi}\: \: \displaystyle \frac{1}{2}< \frac{x}{y}< 2\\ &\textrm{Nilai minimum}\: \: \displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\: \: \textrm{adalah}\: ....\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\color{red}\textrm{Alternatif 1}\\ &\textrm{Misal}\: \: t=\displaystyle \frac{x}{y}\\ &\textrm{Misalkan juga}\: \: f(t)=\displaystyle \frac{x}{2y-x}+\displaystyle \frac{2y}{2x-y}\\ &\textrm{maka}\: \: f(t)=\displaystyle \frac{2t^{2}-3t+4}{-2t^{2}+5t-2}\\ &\bullet \:\textrm{Agar minimum, maka}\: \: f'(t)=0\\ &\: \: \: \: \: \textrm{Sehingga}\\ &\: \: \: \: \: f'(t)=4t^{2}+8t-14=0\\ &\: \: \: \: \Leftrightarrow t_{1,2}=-1\pm \displaystyle \frac{3}{2}\sqrt{2}\\ &\: \: \: \: \textrm{Pilih yang positif, yaitu}\: \: t=-1+ \displaystyle \frac{3}{2}\sqrt{2}\\ &\bullet \: \textrm{Dengan proses substistusi harga}\: \: t\\ &\: \: \: \: \textrm{di atas, maka akan didapatkan }\\ &\: \: \: \: \textrm{nilai}\: \: f(t)=1+\displaystyle \frac{4}{3}\sqrt{2} \end{aligned} \\ &\begin{aligned}&\color{red}\textrm{Alternatif 2}\\ &\textrm{Menurut bentuk}\: \: \displaystyle \frac{1}{2}< \frac{x}{y}< 2\\ &\textrm{jelas bahwa baik}\: \: 2y-x\: \: \textrm{dan}\: \: 2x-y \\ &\textrm{keduanya}\: \textbf{positif}\\ &\color{purple}\textrm{Lihat tabel berikut}\\ &\begin{array}{|c|c|c|}\hline \textrm{Bentuk}&\textrm{Pengecekan 1}&\textrm{Pengecekan 2}\\\hline \begin{aligned}&\displaystyle \frac{1}{2}< \frac{x}{y}< 2\\ &\textrm{Jelas bahwa}\\ &x,y\neq 0\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Saat}\: \: (\times y)\\ &\textrm{Yaitu}:\\ &\displaystyle \frac{1}{2}(y)< \frac{x}{y}(y)< 2(y)\\ &\Leftrightarrow \displaystyle \frac{1}{2}y< x< 2y\\ &\textrm{Jelas bahwa}\\ &2y-x>0\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Saat dibali posisinya}\\ &\displaystyle \frac{1}{2}< \frac{y}{x}< 2\\ &\textrm{Saat}\: \: (\times x)\\ &\textrm{Yaitu}:\\ &\displaystyle \frac{1}{2}(x)< \frac{y}{x}(x)< 2(x)\\ &\Leftrightarrow \displaystyle \frac{1}{2}x< y< 2x\\ &\textrm{Jelas bahwa}\\ &2x-y>0 \end{aligned}\\\hline \end{array}\\ &\textrm{Saat masing-masing}\\ &\bullet \: \displaystyle \frac{x}{2y-x}=\displaystyle \frac{1}{3}+\frac{2}{3}\left ( \displaystyle \frac{2x-y}{2y-x} \right )\: \: \: \textrm{dan}\\ &\bullet \: \displaystyle \frac{2y}{2x-y}=\displaystyle \frac{2}{3}+\frac{4}{3}\left ( \displaystyle \frac{2y-x}{2x-y} \right )\\ &\color{blue}\textrm{Dengan ketaksamaan AM-GM diperoleh}\\ &\begin{aligned} \displaystyle \frac{x}{2y-x}+\displaystyle \frac{2y}{2x-y}&=1+\frac{2}{3}\left ( \displaystyle \frac{2x-y}{2y-x} \right )+\frac{4}{3}\left ( \displaystyle \frac{2y-x}{2x-y} \right )\\ &\geq 1+2\sqrt{\frac{2}{3}\left ( \displaystyle \frac{2x-y}{2y-x} \right )\frac{4}{3}\left ( \displaystyle \frac{2y-x}{2x-y} \right )}\\ &= 1+2\sqrt{\displaystyle \frac{8}{9}}\\ &=1+2\left ( \displaystyle \frac{2}{3} \right )\sqrt{2}\\ &=1+\displaystyle \frac{4}{3}\sqrt{2} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: x,y,z\: \: \textrm{adalah sisi sebuah segitiga}\\ &\textrm{Tunjukkan bahwa nilai minimum}\\ & \displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\: \: \textrm{adalah 3}\\\\ &\textbf{Bukti}\\ &\color{red}\textrm{Alternatif 1}\\ &\begin{aligned}&\textrm{Misalkan}\: \: \displaystyle \frac{y}{x}=p,\: \frac{z}{y}=q,\: \: \textrm{dan}\: \: \displaystyle \frac{z}{x}=pq,\\ &\textrm{maka soal dapat kita ubah menjadi}\\ &\displaystyle \frac{1}{p}+\frac{1}{q}+pq\\ &\textrm{Kita tahu bahwa}\: \: \left ( \sqrt{\displaystyle \frac{1}{p}}-\sqrt{\displaystyle \frac{1}{q}} \right )^{2}\geq 0\\ &\Leftrightarrow \displaystyle \frac{1}{p}+\frac{1}{q}\geq 2\sqrt{\displaystyle \frac{1}{pq}}\\ &\textrm{Kita sesuaikan soal, yaitu}\\ &\displaystyle \frac{1}{p}+\frac{1}{q}+pq\geq pq+2\sqrt{\displaystyle \frac{1}{pq}}\\ &\textrm{Misalkan}\: \: \sqrt{\displaystyle \frac{1}{pq}}=a,\: \displaystyle \frac{1}{pq}=a^{2},\: pq=\displaystyle \frac{1}{a^{2}}\\ &\textrm{dengan}\: \: a\geq 0,\: \: \textrm{maka}\\ &\displaystyle \frac{1}{p}+\frac{1}{q}+pq=\color{blue}\displaystyle \frac{1}{a^{2}}+2a\color{black}=\color{blue}f(a)\\ &\textrm{Syarat ekstrim minimum}\: :\: f'(a)=0\\ &-2a^{-3}+2=0\Leftrightarrow 2=2a^{-3}\Leftrightarrow 1=a^{-3}\\ &\Leftrightarrow 1=\displaystyle \frac{1}{a^{3}}\Leftrightarrow a^{3}=1\Leftrightarrow a=1\\ &\textrm{Sehingga nilai minimumnya saat}\: \: a=1\\ &\textrm{dengan nilai}\: \: f(1)=\displaystyle \frac{1}{1^{2}}+2.1=\color{red}3\\ &\textrm{Jadi, nilai nilai minimum}\: \: \displaystyle \frac{1}{p}+\frac{1}{q}+pq=3\\ &\textrm{atau}\: \: \displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\: \: \textrm{adalah 3 atau juga}\\ &\textrm{dapat dituliskan dengan}\: \: \displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\geq 3 \end{aligned}\\ &\color{red}\textrm{Alternatif 2}\\ &\textrm{Dengan}\: \: \textbf{ketaksamaan AM-GM}\\ &\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\geq 3\sqrt[3]{\displaystyle \frac{x}{y}.\frac{y}{x}.\frac{x}{z}}=3.\sqrt[3]{1}=3 \end{array}$

DAFTAR PUSTAKA

Kartrini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
Widodo, T. 2018. Booklet OSN SMA 2018: Soal dan Solusi OSK, OSP, OSN SMA Bidang Matematika.

SUMBER WEBSITE

Pythagoras pada: https://pyth.eu/uploads/user/ArchiefPDF/Pyth35-56.pdf

Penggunaan Turunan Fungsi (Lanjutan Materi Turunan Fungsi Aljabar)

Penggunaan Turunan Fungsi Aljabar ini nantinya terdapat di antaranya pada:

Persamaan garis singgung
Fungsi naik dan fungsi turun
Menggambar grafik fungsi aljabar
Maksimum dan minimum fungsi
Teorema L'Hopital (dibaca : Lopital)
Titik Stasioner/Titik kritis/Titik Ekstrim (titik maksmum, titik minimum, dan titik belok)
Kecepatan dan percepatan

Perhatikanlah tabel berikut

$\begin{array}{|c|l|l|}\hline \textrm{No}&\quad\textrm{Turunan Pertama}&\qquad\textrm{Turunan Pertama}\\\hline 1.&\textrm{Gradien garis singgung}&\begin{aligned}m&={f}\, '(x)\\&=\underset{h\rightarrow 0}{\textrm{Lim}}=\displaystyle \frac{f(x+h)-f(x)}{h}\end{aligned}\\\hline 2.&\textrm{Fungsi naik dan turun}&y=f(x)\begin{cases} {f}'(x)> 0, & \\ \text{ fungsi naik }&\\\\ {f}'(x)< 0, &\\ \text{ fungsi turun }& \end{cases}\\\hline 3.&\textrm{Jarak, kec, percepatan}&y=s(t)\begin{cases} s(t) & \text{ jarak} \\ {s}\, '(t) & \text{ kecepatan } \\ {s}\, ''(t) & \text{ percepatan}\end{cases}\\\hline4.\textrm{a}&\textrm{Stasioner}&\begin{aligned}\textrm{Maksimum}:&\\ \rightarrow {f}&\, ''(k)< 0\\ \textrm{titik mak}&\: \left ( k, f(k)\right ) \end{aligned}\\\hline 4.\textrm{b}&\textrm{Stasioner}&\begin{aligned}\textrm{Minimum}:&\\ \rightarrow {f}&\, ''(k) > 0\\ \textrm{titik min}&\: \left ( k, f(k)\right ) \end{aligned}\\\hline 4.\textrm{c}&\begin{aligned}&\textrm{Syarat stasioner}\\ &f'(x)=0,\: \: \rightarrow x\\ &\textrm{dengan}\quad x=k \end{aligned}&\begin{aligned}\textrm{Belok}:&\\ \rightarrow {f}&\, ''(k)= 0\\ \textrm{titik belok}&\: \left ( k, f(k)\right ) \end{aligned}\\\hline5.&\begin{aligned}&\textrm{Limit fungsi}\\ &\textrm{bentuk tak tentu}\end{aligned}&\begin{aligned}&\textrm{Aturan L'Hopital}\\ &\\ &\underset{x\rightarrow h}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)}{g(x)}=\underset{x\rightarrow h}{\textrm{Lim}}\: \: \displaystyle \frac{{f}\, '(x)}{g\,'(x)}\\ &\\ &\textrm{untuk hasil limit}\\ &\textrm{bentuk}\: \: \frac{0}{0}\: \: \textrm{atau}\: \: \frac{\infty }{\infty }\end{aligned}\\\hline\textrm{No}&\textrm{Turunan Kedua}&\textrm{Turunan Kedua}\\\hline6.&f''=\displaystyle \frac{d^{2}y}{dx^{2}}&\begin{aligned}\bullet \quad &\textrm{Belok}\\ \bullet \quad &\textrm{Percepatan}\\ \bullet \quad &\textrm{Maksimum}\\ \bullet \quad &\textrm{Minimum}\end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah persamaan garis singgung }\\ &\textrm{pada kurva}\: \: y=x^{2}+2x-8\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: 1\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahui},\: \textrm{persamaan sebuah kurva adalah:}\\ &y=\color{blue}x^{2}+2x-8\\ & \end{aligned}\\ &\begin{array}{|c|c|}\hline \textrm{Titik singgung}&\textrm{Gradien}\: ,\: x=1\\\hline \begin{aligned}\textrm{absis}\: \: x&=1,\\ \textrm{maka}\: \: y&=(1)^{2}+2(1)-8\\ &=1+2-8\\ &=-5\\ &\\ \textrm{di titik}&\: \: (a,b)=(1,-5)\end{aligned}&\begin{aligned}\displaystyle \frac{dy}{dx}=m&=2x+2\\ &=2(1)+2\\ &=4\\ &\\ & \end{aligned}\\\hline \textrm{Persamaan garis singgung}&\textrm{Kesimpulan}\\\hline\begin{aligned}y&=m(x-a)+b\\ &=4(x-1)+(-5)\\ &=4x-4-5\\ &=4x-9\end{aligned}&\begin{aligned}&\textrm{Sehingga},\: \textrm{PGS adalah:}\\ &y=\color{red}4x-9\qquad \color{black}\textbf{atau}\qquad\\ &\color{red}y-4x+9=0\\ & \end{aligned} \\\hline \end{array}\\ &\begin{aligned}&\textrm{Jadi},\: \textrm{persamaan garis singgungnya adalah:}\\ &y=\color{red}4x-9\qquad \color{black}\textbf{atau}\qquad \color{red}y-4x+9=0\\\\ &\textrm{Dan berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah interval di mana kurva }\\ &\textrm{fungsi}\: \: f(x)=x^{3}+3x^{2}-9x+5\\ &\textrm{a}.\quad \textrm{naik}\\ &\textrm{b}.\quad \textrm{turun}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui fungsi}\\f(x)&=x^{3}+3x^{2}-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ &=3(x+3)(x-1)\\ & \end{aligned}\\ &\begin{array}{|l|l|}\hline \qquad\textrm{naik}\: ;\: \: {f}\, '(x)> 0&\quad\textrm{turun}\: ;\: \: {f}\, '(x)< 0\\\hline \begin{aligned}&\\ &3(x+3)(x-1)> 0\\ & \end{aligned}&\begin{aligned}&\\ &3(x+3)(x-1) < 0\\ & \end{aligned}\\\hline \begin{aligned}&\\ \textrm{naik}\: ,&\: x< -3\: \: \textrm{atau}\: \: x> 1\\ & \end{aligned}&\begin{aligned}&\\ \textrm{turun}\: ,&\: -3< x< 1\\ & \end{aligned}\\\hline \end{array} \end{array}$ .

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah nilai stasioner fungsi}\\ & f(x)=x^{3}+3x^{2}-9x+5 \: \: \textrm{dan }\\ &\textrm{tentukan pula jenisnya}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahu fungsi}\\&f(x)=x^{3}+3x^{2}-9x+5\\ &\color{blue}\textrm{Syarat stasioner}\: \: \color{black}f'(x)=0,\: \: \textrm{maka}\\ &{f}\, '(x)=3x^{2}+6x-9=0\\ &\qquad 0=3(x+3)(x-1)\\ &\: \: \: \qquad x=-3\: \: \textrm{atau}\: \: x=1\\ &\color{blue}\textrm{Dan untuk nilai dan titik stasionernya}: \\ &f(-3)=(-3)^{3}+3(-3)^{2}-9(-3)+5=32\\ &\rightarrow \left ( -3,32 \right )\: \color{red}\textrm{adalah titik balik maksimum}\\\\ &f(1)=(1)^{3}+3(1)^{2}-9(1)+5=0\\ &\rightarrow \left ( 1,0 \right )\: \color{red}\textrm{adalah titik balik minimum}\\\\ &\textrm{Dan berikut ilustrasi gambarnya}\\ &\textrm{untuk soal no.2 dan 3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Masih sama dengan soal seperti pada }\\ &\textrm{No. sebelumnya yaitu fungsi}\\ &f(x)=x^3+3x^2-9x+5\: .\\ & \textrm{Tentukanlah koordinat titik beloknya}\\\\ &\textbf{Jawab}:\\ &\begin{aligned} f(x)&=x^3+3x^2-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ {f}\, ''(x)&=6x+6\\ \textrm{Proses}\: &\textrm{mencari titik beloknya}\\ {f}\, ''(x)&=0\\ 6x+6&=0\\ 6x&=-6\\ x&=-1\\ & \end{aligned} \\ &\begin{array}{|c|c|c|c|l|}\hline \textrm{Interval}&f(x)&{f}\, '(x)&{f}\, ''(x)&\: \: \qquad \textrm{Keterangan}\\\hline x< -1&&&-&\textrm{grafik cekung ke bawah}\\\hline x= -1&16&-12&0&\textrm{grafik memiliki titik belok}\\\hline x > -1&&&+&\textrm{grafik cekung ke atas}\\\hline \end{array} \\ &\textrm{Koordinat titik beloknya}: \: (-1,16)\\\\ &\textrm{Berikut ilustrasi gambarnya} \end{array}$.

Aturan Rantai pada Turunan Pertama dan Turunan Kedua Fungsi Aljabar (Lanjutan Materi Turunan Fungsi Aljabar)

Aturan Rantai

$\begin{aligned}&\\ \textrm{Untuk}:&\\ &\begin{cases} a & \in \mathbb{R} \\ n& \in \mathbb{Q}\\ c& \textrm{konstanta}\\ U&=g(x)\\ V&=h(x) \end{cases}\\ & \end{aligned}$.

$\begin{array}{|l|}\hline \color{red}\textbf{Sifat-Sifat}\\\hline \begin{aligned}y&=c\rightarrow \qquad{y}'=0\\ y&=c.U\rightarrow \quad{y}'=c.{U}'\\ y&=U\pm V\rightarrow {y}'={U}'\pm {V}'\\ y&=U.V\rightarrow \: \: \: \: {y}'={U}'.V+U.{V}'\\ y&=\displaystyle \frac{U}{V}\rightarrow \quad\: \: \: {y}'=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}\\\hline \color{red}\textbf{Fungsi Aljabar}\\\hline \begin{aligned}y&=a.x^{n}\rightarrow {y}\: '=n.a.x^{(n-1)}\\ y&=a.U^{n}\rightarrow {y}\: '=n.a.U^{(n-1)}.{U}'\end{aligned}\\\hline \color{red}\textbf{Fungsi Trigonometri}\\\hline \begin{aligned}y&=a\sin U\rightarrow & {y}'&=\left (a\cos U \right ).{U}'\\ y&=a\cos U\rightarrow &{y}'&=\left (-a\sin U \right ).{U}'\\ y&=a\tan U\rightarrow &{y}'&=\left (a\sec ^{2}U \right ).{U}'\end{aligned}\\\hline \color{red}\textbf{Aturan rantai }\\\hline \begin{aligned} &\textrm{pada turunan untuk} \: \: y=f(u),\: \: \textrm{jika}\\ &\textrm{untuk}\: \: u\: \: \textrm{merupakan fungsi}\: \: x,\: \: \textrm{maka}:\\ &\color{blue}{y}\: '={f}\: '(x).{u}'\: \: \color{black}\textrm{atau}\: \: \: \: \color{blue}\displaystyle \frac{dy}{dx}=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ & \end{aligned} \\\hline \end{array}$.

Turunan Kedua Fungsi Aljabar

$\begin{aligned}&\\ \textrm{Suatu fungsi}\qquad &\\ &f\: :\: x\: \rightarrow \: y\\ & \end{aligned}$.

$\begin{array}{|c|c|}\hline \textrm{Notasi}&\textrm{Proses}\\\hline \begin{aligned}&\\ &\begin{cases} {f}\: ' (x)& =\displaystyle \frac{dy}{dx} \\\\ {f}\: ''(x) & =\displaystyle \frac{d^{2}y}{dx^{2}} \end{cases}\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ y&=ax^{n}\\ {y}'&=n.a.x^{n-1}\\ {y}''&=(n-1).n.a.x^{n-2}\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai dari}\: \: \: \displaystyle \frac{dy}{dx}\\ &\textrm{a}.\quad y=\sqrt{x}\\ &\textrm{b}.\quad y=\sqrt{3+\sqrt{x}}\\ &\textrm{c}.\quad y=\sqrt{3+\sqrt{3+\sqrt{x}}}\\ &\textrm{d}.\quad y=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{x}}}}\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline \begin{aligned}\textrm{a}.\qquad y&=\sqrt{x}\\ &=x^{^{\frac{1}{2}}}\\ {y}'&=\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2x^{^{\frac{1}{2}}}}\\ &=\displaystyle \frac{1}{2\sqrt{x}}\\ &\\ &\\ & \end{aligned}&\begin{aligned} \textrm{b}.\qquad y&=\sqrt{3+\sqrt{x}}\\ &=\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}}}\\ {y}'&=\displaystyle \frac{1}{2}\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}-1}}.\displaystyle \frac{1}{2}x^{^{\frac{1}{2}-1}}\\ &=\displaystyle \frac{1}{4}\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{-\frac{1}{2}}}.x^{^{-\frac{1}{2}}}\\ &=\displaystyle \frac{1}{4\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}}}.x^{^{\frac{1}{2}}}}\\ &=\displaystyle \frac{1}{4\sqrt{3+\sqrt{x}}.\sqrt{x}}\\ & \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=2\sin x\cos x&\textrm{k}.&f(x)=4\sin \left ( 5x+\pi \right )^{3}\\ \textrm{b}.&f(x)=x^{2}.\sin x&\textrm{l}.&f(x)=\cos ^{3}(x+5)&\\ \textrm{c}.&f(x)=3\cos ^{2}x&\textrm{m}.&f(x)=\sin ^{2}\left ( \pi -3x \right )&\\ \textrm{d}.&f(x)=3\sin ^{2}x-x^{3}&\textrm{n}.&f(x)=\displaystyle \frac{\sin ^{2}x}{\cos x}&\\ \textrm{e}.&f(x)=5\sin ^{4}x&\textrm{o}.&f(x)=\displaystyle \frac{\sin x^{4}}{x^{2}}&\\ \textrm{f}.&f(x)=\sin \left ( \displaystyle \frac{x+3}{x-2} \right )\\ \textrm{g}.&f(x)=\cos \left ( x^{2}-6x \right )\\ \textrm{h}.&f(x)=\displaystyle \frac{1}{\sin x}\\ \textrm{i}.&f(x)=\displaystyle \frac{\sin x}{1+\cos x}\\ \textrm{j}.&f(x)=\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x} \end{array} \end{array}$.

$\begin{aligned}&\textbf{Jawab}:\\ &\begin{array}{ll}\\ 2.\: \textrm{a}&\color{blue}\textrm{Alternatif 1}\\ &f(x)=2\sin x\cos x\\ &\begin{aligned}f(x)&=\sin 2x\\ {f}\: '(x)&=\cos 2x.(2)\\ &=\color{red}2\cos 2x \end{aligned}\\\\ &\color{blue}\textrm{Alternatif 2}\\ &f(x)=y=\sin 2x\\ &\begin{cases} y=\sin u & \rightarrow \displaystyle \frac{dy}{du}=\cos u\\\\ u=2x & \rightarrow \displaystyle \frac{du}{dx}=2 \end{cases} \\ &\textrm{sehingga}\\ &\begin{aligned}y&=\sin 2x\quad \\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ &=\cos u. 2\\ &=2\cos u\\ &=2\cos 2x \end{aligned} \end{array} \end{aligned}$.

$\begin{array}{ll}\\ 2.\: \textrm{b}&f(x)=x^{2}\sin x\\ &\begin{aligned} {f}\: '(x)&=2x^{2-1}.(\sin x)+x^{2}.(\cos x)\\ &=2x\sin x+x^{2}\cos x \end{aligned}\\ \: \: \: \: \textrm{c}&f(x)=3\cos ^{2}x\\ &f'(x)=2.3.\cos^{2-1} x.(-\sin x)\\ &\: \: \qquad =-6\sin x\cos x\\ &\: \: \qquad =-3\sin 2x\\ \: \: \: \: \textrm{d}&f(x)=3\sin ^{2}x-x^{3}\\ &f'(x)=2.3.\sin^{2-1} x.(\cos x)-3x^{3-1}\\ &\: \: \qquad =6\sin x\cos x-3x^{2}\\ &\: \: \qquad =3\sin 2x-3x^{2}\\ \: \: \: \: \textrm{e}&f(x)=5\sin ^{4}x\\ &f'(x)=4.5.\sin^{4-1} x.(\cos x)\\ &\: \: \qquad =20\sin^{3} x\cos x\\ &\: \: \qquad \color{red}\textrm{atau boleh juga}\\ &\: \: \qquad =20\sin^{2} x\sin x\cos x\\ &\: \: \qquad =20\sin^{2} x\sin 2x=20\sin 2x\sin ^{2}x\\ \end{array}$ .

$\begin{array}{ll}\\ 2.\textrm{m}&y=f(x)=\sin ^{2}(\pi -3x)\\ &\begin{aligned}\quad{f}\: '(x)&=2\sin ^{2-1}\left ( \pi -3x \right ).\cos \left ( \pi -3x \right ).(-3)\\ &=-6\sin \left ( \pi -3x \right )\cos \left ( \pi -3x \right )\\ &=-3.2\sin \left ( \pi -3x \right )\cos \left ( \pi -3x \right )\\ &=-3\sin 2\left ( \pi -3x \right )\\ &=-3\sin \left ( 2\pi -6x \right ) \end{aligned}\\\\ &\color{red}\textrm{atau boleh juga}\\ &\begin{aligned}y=&f(x)=\sin ^{2}\left ( \pi -3x \right )\\ &\quad \begin{cases} y=u^{2} &\rightarrow \displaystyle \frac{dy}{du}=2u. \\\\ u=\sin w &\rightarrow \displaystyle \frac{du}{dw}=\cos w \\\\ w=\pi -3x &\rightarrow \displaystyle \frac{dw}{dx}=-3 \end{cases}\\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dw}.\frac{dw}{dx}\\ &=\left ( 2u \right ).\left ( \cos w \right ).\left ( -3 \right )\\ &= -3.\left ( 2\sin w \right ).\left ( \cos w \right )\\ &=-3\sin 2w\\ &=-3\sin \left ( 2\pi -6x \right )\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah nilai dari}\: \: \: \displaystyle \frac{d^{2}y}{dx^{2}}\\ &\textrm{a}.\quad y=\sqrt{x}\\ &\textrm{b}.\quad y=\sqrt{3+\sqrt{x}}\\ &\textrm{c}.\quad y=15x^{3}-4x\\ &\textrm{d}.\quad y=\displaystyle \frac{x^{4}}{3}+\frac{x^{2}}{2}+x+\sqrt{x}+1+\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^{2}}+\frac{3}{x^{4}}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Lihat pembahasan soal no.1, yaitu }\\ &y=\sqrt{x}\\ &y'=\displaystyle \frac{1}{2\sqrt{x}}\end{aligned}\\ &\textrm{maka}\\ &\begin{aligned}3.\textrm{a}.\quad y&=\sqrt{x}\\ y'&=\displaystyle \frac{dy}{dx}=\displaystyle \frac{1}{2\sqrt{x}}=\displaystyle \frac{1}{2}x^{-\frac{1}{2}}\\ y''&=\displaystyle \frac{d^{2}y}{dx^{2}}=\left ( -\displaystyle \frac{1}{2} \right )\displaystyle \frac{1}{2}.x^{-\frac{1}{2}-1}\\ &=-\displaystyle \frac{1}{4x^{\frac{1}{2}+1}}=-\displaystyle \frac{1}{4x\sqrt{x}} \end{aligned} \end{array}$.

$.\qquad\: \begin{array}{|l|}\hline \begin{aligned}&3.\textrm{d}\\ &\begin{aligned}y&=\displaystyle \frac{x^{4}}{3}+\frac{x^{2}}{2}+x+\sqrt{x}+1+\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^{2}}+\frac{3}{x^{4}}\\ &\qquad\quad \parallel \\ &\textbf{Turunan pertama}\\ &\qquad\quad \Downarrow \\ \displaystyle \frac{dy}{dx}={y}\, '&=\displaystyle \frac{4x^{3}}{3}+x+1+\displaystyle \frac{1}{2\sqrt{x}}+0-\frac{1}{2x\sqrt{x}}-\frac{1}{x^{2}}-\frac{2}{x^{3}}-\frac{12}{x^{5}}\\ &=\displaystyle \frac{4x^{3}}{3}+x+1+\displaystyle \frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}-\frac{1}{x^{2}}-\frac{2}{x^{3}}-\frac{12}{x^{5}}\\ &\qquad\quad \parallel \\ &\textbf{Turunan kedua}\\ &\qquad\quad \Downarrow \\ \displaystyle \frac{d^{2}y}{dx^{2}}={y}\, ''&=4x^{2}+1+0-\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^{2}\sqrt{x}}+\frac{2}{x^{3}}+\frac{6}{x^{4}}+\frac{60}{x^{6}}\\ &=4x^{2}+1-\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^{2}\sqrt{x}}+\frac{2}{x^{3}}+\frac{6}{x^{4}}+\frac{60}{x^{6}} \end{aligned}\\ & \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

Silahkan kerjakan soal yang belum diselesaikan atau dijawab

DAFTAR PUSTAKA

Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU
Kartrini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: ESIS
Sunardi, Waluyo, S., Sutrisno, Subagya. 2004. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.

Sifat Turunan Pertama dan Aturan Rantai pada Turunan Fungsi Aljabar

Rumus Turunan dan Sifat Turunan Pertama

$\begin{aligned}&\\ \textrm{Untuk}:&\\ &\begin{cases} a & \in \mathbb{R} \\ n& \in \mathbb{Q}\\ c& \textrm{konstanta}\\ U&=g(x)\\ V&=h(x) \end{cases}\\ & \end{aligned}$.

$\begin{array}{|l|}\hline \color{blue}\textbf{Sifat-Sifat}\\\hline \begin{aligned}y&=c\rightarrow \qquad{y}'=0\\ y&=c.U\rightarrow \quad{y}'=c.{U}'\\ y&=U\pm V\rightarrow {y}'={U}'\pm {V}'\\ y&=U.V\rightarrow \: \: \: \: {y}'={U}'.V+U.{V}'\\ y&=\displaystyle \frac{U}{V}\rightarrow \quad\: \: \: {y}'=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}\\\hline \color{blue}\textbf{Fungsi Aljabar}\\\hline \begin{aligned}y&=a.x^{n}\rightarrow {y}\: '=n.a.x^{(n-1)}\\ y&=a.U^{n}\rightarrow {y}\: '=n.a.U^{(n-1)}.{U}'\end{aligned}\\\hline \color{blue}\textbf{Fungsi Trigonometri}\\\hline \begin{aligned}y&=a\sin U\rightarrow & {y}'&=\left (a\cos U \right ).{U}'\\ y&=a\cos U\rightarrow &{y}'&=\left (-a\sin U \right ).{U}'\\ y&=a\tan U\rightarrow &{y}'&=\left (a\sec ^{2}U \right ).{U}'\end{aligned}\\\hline \color{blue}\textbf{Aturan rantai }\\\hline \begin{aligned} &\textrm{pada turunan untuk} \: \: y=f(u),\: \: \textrm{jika}\\ &\textrm{untuk}\: \: u\: \: \textrm{merupakan fungsi}\: \: x,\: \: \textrm{maka}:\\ &\color{red}{y}\: '={f}\: '(x).{u}'\: \: \color{black}\textrm{atau}\: \: \: \: \color{red}\displaystyle \frac{dy}{dx}=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ & \end{aligned} \\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Dengan menggunakan rumus turunan}\\ &{f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h},\\ &\textrm{tunjukkan bahwa turunan}\: \: \\ &\textrm{a}.\quad f(x)=ax^{n}\: \: \textrm{adalah}\: \: {f}\, '(x)=n.ax^{n-1}\\ &\textrm{b}.\quad f(x)=u(x)+v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x)+{v}\,'(x)\\ &\textrm{c}.\quad f(x)=u(x).v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x).v(x)+u(x).{v}\,'(x)\\ &\textrm{d}.\quad f(x)=\displaystyle \frac{u(x)}{v(x)}\: \: \textrm{adalah}\: \: {f}\, '(x)=\displaystyle \frac{{u}\,'(x).v(x)+u(x).{v}\,'(x)}{(v(x))^{2}} \\ &\textrm{e}.\quad f(x)=\sin x\: \: \textrm{adalah}\: \: {f}\, '(x)=\cos x \\ &\textrm{f}.\quad f(x)=\cos x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\sin x \\ &\textrm{g}.\quad f(x)=\tan x\: \: \textrm{adalah}\: \: {f}\, '(x)=\sec^{2} x \\ &\textrm{h}.\quad f(x)=\cot x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\csc^{2} x \\ \end{array}$.

$\begin{aligned}&\textbf{Bukti}\\ &\begin{array}{ll}\\ 1.\: \textrm{a}&{f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\, '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a(x+h)^{n}-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left ( x^{n}+\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left (\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\left (\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}.h+\binom{n}{3}x^{n-3}.h^{2}+\cdots +\binom{n}{n-1}x.h^{n-2}+h^{n-1}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle a\binom{n}{1}x^{n-1}+a\binom{n}{2}x^{n-2}.h+a\binom{n}{3}x^{n-3}.h^{2}+\cdots +ah^{n-1}\\ &=a\binom{n}{1}x^{n-1}+0+0+...+0\\ &=a.\displaystyle \frac{n!}{(n-1)!.1!}x^{n-1}\\ &=a.\displaystyle \frac{n.(n-1)!}{(n-1)!}x^{n-1}\\ &=a.n.x^{n-1}\qquad\quad \blacksquare \end{aligned} \end{array} \end{aligned}$.

$\begin{array}{ll}\\ 1.\: \textrm{b}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (u(x+h)+v(x+h) \right )-\left ( u(x)+v(x) \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \left (\displaystyle \frac{u(x+h)-u(x)}{h}+\frac{v(x+h)-v(x)}{h} \right )\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{u(x+h)-u(x)}{h}+ \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{v(x+h)-v(x)}{h}\\ &=u'(x)+v'(x)\qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 1.\: \textrm{c}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (u(x+h)\times v(x+h) \right )-\left ( u(x)\times v(x) \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{u(x+h)\times v(x+h)-u(x+h)\times v(x)+u(x+h)\times v(x)-u(x)\times v(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \left (u(x+h)\times \displaystyle \frac{v(x+h)-v(x)}{h}+v(x)\times \displaystyle \frac{u(x+h)-u(x)}{h} \right )\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: u(x+h)\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{v(x+h)-v(x)}{h}+\underset{h\rightarrow 0}{\textrm{Lim}}\: \: v(x)\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{u(x+h)-u(x)}{h}\\ &=u(x)\times v'(x)+v(x)\times u'(x)\\ &=u'(x)\times v(x)+u(x)\times v'(x)\qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 1.\: \textrm{d}&\textrm{Misalkan}\: \: p(x)=\displaystyle \frac{u(x)}{v(x)}\\ &\textrm{Sebelumnya telah diketahui dari no. 1. c}\\ &\: {u}\: (x)=p(x)\times v(x)\\ &\begin{aligned}{u}\: '(x)&=p'(x)\times v(x)+p(x)\times v'(x)\\ \textrm{Sekar}&\textrm{ang kita substitusikan pemisalan}\\ \textrm{di at}&\textrm{as, yaitu}: \end{aligned}\\ &\begin{aligned}\: p'(x)&\times v(x)=u'(x)-p(x)\times v'(x)\\ &=u'(x)-\displaystyle \frac{u(x)}{v(x)}\times v'(x)\\ &=\displaystyle \frac{u'(x)\times v(x)-u(x)\times v'(x)}{v(x)}\\ p'(x)&=\displaystyle \frac{u'(x)\times v(x)-u(x)\times v'(x)}{v^{2}(x)}\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 1.\: \textrm{e}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)-\sin x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2\cos \displaystyle \frac{1}{2}(2x+h)\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 2\cos \displaystyle \frac{1}{2}(2x+h).\displaystyle \frac{\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 2\cos \displaystyle \frac{1}{2}(2x+h)\times \displaystyle \frac{1}{2}\\ &=2\cos \displaystyle \frac{1}{2}(2x+0)\times \displaystyle \frac{1}{2}\\ &=\cos \displaystyle \frac{1}{2}(2x)\\ &=\cos x\qquad\quad \blacksquare \end{aligned} \end{array}$ .

$\begin{array}{ll}\\ 1.\: \textrm{f}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\cos (x+h)-\cos x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2\sin \displaystyle \frac{1}{2}(2x+h)\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle -2\sin \displaystyle \frac{1}{2}(2x+h).\frac{\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle -2\sin \displaystyle \frac{1}{2}(2x+h)\times \frac{1}{2}\\ &=-2\sin \displaystyle \frac{1}{2}(2x+0)\times \frac{1}{2}\\ &=-\sin \displaystyle \frac{1}{2}(2x)\\ &=-\sin x\qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 1.\: \textrm{g}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h}{1-\tan x.\tan h}-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h-\tan x+\tan^{2}x.\tan h}{1-\tan x.\tan h}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h\left ( 1+\tan^{2}x \right )}{h\left ( 1-\tan x.\tan h \right )}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h}{h}\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{1+\tan^{2}x}{1-\tan x.\tan h}\\ &=1 \times \displaystyle \frac{1+\tan^{2}x}{1-0}\\ &=1+\tan^{2}x\\ &=\sec ^{2}x\qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=x^{6}&\textrm{i}.&f(x)=(x+1)(x-2)&\textrm{q}.&f(x)=\displaystyle \frac{2}{x^{3}}\\ \textrm{b}.&f(x)=\displaystyle \frac{1}{2}x^{2}&\textrm{j}.&f(x)=(3-x)(5-x)&\textrm{r}.&f(x)=\displaystyle \frac{1}{2\sqrt{x}}\\ \textrm{c}.&f(x)=-2x^{5}&\textrm{k}.&f(x)=(2x+3)^{2}&\textrm{s}.&f(x)=\displaystyle \frac{1}{3x^{5}}\\ \textrm{d}.&f(x)=ax^{3}&\textrm{l}.&f(x)=(x-2)^{3}&\textrm{t}.&f(x)=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ \textrm{e}.&f(x)=4x^{4}-x^{2}+2017&\textrm{m}.&f(x)=(4x+1)(4x-1)&\textrm{u}.&f(x)=\displaystyle \frac{x}{2}+\frac{2}{x}\\ \textrm{f}.&f(x)=6-x-3x^{2}&\textrm{n}.&f(x)=(x-1)(x+1)(x+2)&\textrm{v}.&f(x)=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ \textrm{g}.&f(x)=(x-3)^{2}&\textrm{o}.&f(x)=\displaystyle \frac{1}{2}x^{-4}&\textrm{w}.&f(x)=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ \textrm{h}.&f(x)=\left ( x^{3}-2 \right )^{2}&\textrm{p}.&f(x)=x^{-5}&\textrm{x}.&f(x)=\displaystyle \frac{2+4x}{x^{2}}\\ &&&&\textrm{y}.&f(x)=\left ( x+1+\displaystyle \frac{1}{x} \right )\left ( x+1-\displaystyle \frac{1}{x} \right ) \end{array} \end{array}$.

$\begin{aligned}&\textrm{Untuk}\: \: y=f(x)\: \: \textrm{maka}\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned}y&=x^{6}\\ {y}\: '&=6x^{6-1}\\ &=6x^{5}\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{2}\\ {y}\: '&=2.\displaystyle \frac{1}{2}x^{2-1}\\ &=x \end{aligned}&\begin{aligned}y&=-2x^{5}\\ {y}\: '&=-5.2x^{5-1}\\ &=-10x^{4}\\ & \end{aligned}\\\hline \textrm{d}&\textrm{e}&\textrm{f}\\\hline \begin{aligned}y&=ax^{3}\\ {y}\: '&=3.a.x^{3-1}\\ &=3.a.x^{2} \end{aligned}&\begin{aligned}y&=4x^{4}-x^{2}+2017\\ {y}\: '&=4.4x^{4-1}-2.x^{2-1}+0\\ &=16x^{3}-2x \end{aligned}&\begin{aligned}y&=6-x-3x^{2}\\ {y}\: '&=0-1.x^{1-1}-2.3x^{2-1}\\ &=-1-6x \end{aligned}\\\hline \end{array} \end{aligned}$.

$\begin{array}{|c|c|}\hline \textrm{g}&\textrm{h}\\\hline \begin{aligned}y&=(x-3)^{2}\\ {y}\: '&=2.(x-3)^{2-1}.1\\ &=2(x-3)\\ &\end{aligned}&\begin{aligned}y&=\left ( x^{3}-2 \right )^{2}\\ {y}\: '&=2.\left ( x^{3}-2 \right )^{2-1}.3x^{2}\\ &=6\left ( x^{3}-2 \right )x^{2}\\ &=6x^{5}-12x^{2} \end{aligned}\\\hline \textrm{i}&\textrm{j}\\\hline \begin{aligned}y&=(x+1)(x-2)\\ {y}\: '&=1.(x+2)+(x+1).1\\ &=2x+3 \end{aligned}&\begin{aligned}y&=(3-x)(5-x)\\ {y}\: '&=-1.(5-x)+(3-x).-1\\ &=2x-8 \end{aligned}\\\hline \end{array}$ .

$\begin{array}{|c|c|c|c|}\hline \textrm{k}&\textrm{l}&\textrm{m}_{1}&\textrm{m}_{2}\\\hline \begin{aligned}y&=(2x+3)^{2}\\ {y}\: '&=2.(2x+3)^{2-1}.2\\ &=4(2x+3)^{1}\\ &=8x+12\\ &\\ & \end{aligned}&\begin{aligned}y&=(x-2)^{3}\\ {y}\: '&=3(x-2)^{3-1}.1\\ &=3(x-2)^{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 1}\\ {y}\: '&=4(4x-1)+(4x+1).4\\ &=16x-4+16x+4\\ &=32x\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 2}\\ y&=16x^{2}-1\\ {y}\: '&=2.16x^{2-1}-0\\ &=32x^{1}\\ &=32x \end{aligned}\\\hline \textrm{n}&\textrm{o}&\textrm{p}&\textrm{q}\\\hline \begin{aligned}y&=(x-1)(x+1)(x+2)\\ &=(x^{2}-1)(x+2)\\ &=x^{3}+2x^{2}-x-2\\ {y}\: '&=3x^{3-1}+2.2x^{2-1}-x^{1-1}-0\\ &=3x^{2}+4x-1\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{-4}\\ {y}\: '&=-4.\displaystyle \frac{1}{2}x^{-4-1}\\ &=-2x^{-5}\\ &=-\displaystyle \frac{2}{x^{5}}\\ & \end{aligned}&\begin{aligned}y&=x^{-5}\\ {y}'\: &=-5.x^{-5-1}\\ &=-5x^{-6}\\ &=-\displaystyle \frac{5}{x^{6}}\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2}{x^{3}}\\ &=2x^{-3}\\ {y}\: '&=-3.2x^{-3-1}\\ &=-6x^{-4}\\ &=-\displaystyle \frac{6}{x^{4}} \end{aligned}\\\hline \end{array}$.

$\begin{array}{|c|c|c|c|}\hline \textrm{r}&\textrm{s}&\textrm{t}&\textrm{u}\\\hline \begin{aligned}y&=\displaystyle \frac{1}{2\sqrt{x}}\\ &=\displaystyle \frac{1}{2x^{\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ {y}\: '&=-\displaystyle \frac{1}{2}.\frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{4}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{4x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{4}\sqrt{x^{3}} \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{3x^{5}}\\ &=\displaystyle \frac{1}{3}x^{-5}\\ {y}\: '&=-5.\displaystyle \frac{1}{3}x^{-5-1}\\ &=-\displaystyle \frac{5}{3}x^{-6}\\ &=-\displaystyle \frac{5}{3x^{6}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ &=2x^{4}+\displaystyle \frac{1}{2}x^{-3}\\ {y}\: '&=4.2x^{4-1}+(-3).\displaystyle \frac{1}{2}x^{-3-1}\\ &=8x^{3}-\displaystyle \frac{3}{2}x^{-4}\\ &=8x^{3}-\displaystyle \frac{3}{2x^{4}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{x}{2}+\frac{2}{x}\\ &=\displaystyle \frac{1}{2}x+2x^{-1}\\ {y}\: '&=\displaystyle \frac{1}{2}x^{1-1}+(-1).2x^{-1-1}\\ &=\displaystyle \frac{1}{2}x^{0}-2x^{-2}\\ &=\displaystyle \frac{1}{2}-\displaystyle \frac{2}{x^{2}}\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$.

$\begin{array}{|c|c|c|c|}\hline \textrm{v}&\textrm{w}\\\hline \begin{aligned}y&=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ &=\left ( 2x^{3}+x^{-1} \right )^{2}\\ {y}\: '&=2.\left ( 2x^{3}+x^{-1} \right )^{2-1}.\left ( 3.2x^{3-1}+(-1)x^{-1-1} \right )\\ &=2.\left ( 2x^{3}+x^{-1} \right )^{1}.\left ( 6x^{2}-x^{-2} \right )\\ &=2\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )\left ( 6x^{2}-\displaystyle \frac{1}{x^{2}} \right )\\ &=2\left ( 12x^{5}-2x+6x-\displaystyle \frac{1}{x^{3}} \right )\\ &=24x^{5}+8x-\displaystyle \frac{2}{x^{3}}\\ & \end{aligned}&\begin{aligned}y&=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ &=x^{2}\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}\\ {y}\: '&=2x.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}+x^{2}.2.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2-1}.\left ( 0+\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( 1+\sqrt{x} \right ).\left ( \displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( \displaystyle \frac{1}{2x^{^{\frac{1}{2}}}} \right ).\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{2-\frac{1}{2}}}.\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{\frac{3}{2}}}\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+\left ( x\sqrt{x}+x^{2} \right ) \end{aligned}\\\hline \end{array}$.

$\begin{array}{|c|c|}\hline \textrm{x}_{1}&\textrm{x}_{2}\\\hline \begin{aligned}y&=\displaystyle \frac{U}{V}\\ {y}\: '&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}&\begin{aligned}y&=UV\\ {y}\: '&={U}'.V+U.{V}' \end{aligned}\\\hline \begin{aligned}&&&\\ U&=2+4x&\rightarrow {U}'&=4\\ V&=x^{2}&\rightarrow {V}'&=2x\\ &&& \end{aligned}&\begin{aligned}&&&\\ U&=2+4x&\rightarrow {U}'&=4\\ V&=x^{2}&\rightarrow {V}'&=2x\\ &&& \end{aligned}\\\hline \begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ {y}\: '&=\displaystyle \frac{(4)(x^{2})-(2+4x).(2x)}{\left ( x^{2} \right )^{2}}\\ &= \displaystyle \frac{4x^{2}-4x-8x^{2}}{x^{4}}\\ &=\displaystyle \frac{-4x^{2}-4x}{x^{4}}\\ &=\displaystyle \frac{-4x-4}{x^{3}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ &=(2+4x).x^{-2}\\ {y}\: '&=(4).x^{-2}+(2+4x).-2x^{-2-1}\\ &=4x^{-2}-(4+8x).x^{-3}\\ &=4x^{-2}-4x^{-3}-8x^{-2}\\ &=-4x^{-3}-4x^{-2}\\ &=\displaystyle \frac{-4}{x^{3}}+\displaystyle \frac{-4}{x^{2}}\\ &=\displaystyle \frac{-4-4x}{x^{3}}\\ &=\displaystyle \frac{-4x-4}{x^{3}} \end{aligned}\\\hline \end{array}$.

Pengertian dan Bentuk Umum Turunan Fungsi Aljabar (Materi Lanjutan Turunan Fungsi Aljabar)

A. 2 Pengertian Turunan Fungsi Aljabar

Perhatikan ilustrasi gambar berikut.

Misalkan diketahui fungsi $y=f(x)$ terdefinisi pada semua nilai $x$ di sekitar $x=k$. Jika $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ ada, maka bentuk $\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$ disebut sebagai turunan dari fungsi $f(x)$ saat $x=k$.

A. 3 Notasi

Notasi turunan fungsi dilambangkan dengan $f'(k)$ dengan $f'(k)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}$.
Lambang $f'(k)$ dibaca $f$ aksen $k$ disebut turunan atau derivatif untuk fungsi $f(x)$ terhadap $x$ saat $x=k$.
Jika limitnya ada, dapat dikatakan fungsi $f(x)$ diferensiabel (dapat dideferensialkan) saat $x=k$ dan bentuk limitnya selanjutnya dilambangkan dengan $f'(k)$.
Misalkan fungsi $f(x)$ mempunyai turunan $f'(x)$. Jika $f'(k)$ tidak terdefinisi, maka $f(x)$ tidak diferensiabel di $x=k$.

A. 4 Bentuk Umum Turunan Pertama Fungsi Aljabar

Bentuk umum turunan pertama fungsi aljabar untuk fungsi $y$ terhadap $x$ dinotasikan sebagaimana berikut

${y}\: '=\displaystyle \frac{dy}{dx}=\displaystyle \frac{d\left ( f(x) \right )}{dx}={f}'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: f(x)=2x, \textrm{hitunglah laju }\\ &\textrm{perubahan fungsi}\: \: f\: \: \textrm{di}\: \: x=2\\ &\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: \color{blue}f(x)=2x\\ &\begin{array}{|l|}\hline \color{blue}\textrm{Cara Pertama}\\\hline\begin{aligned} f(x)&=2x\\ f(2)&=2.2=4\\ {f}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(2x)-(4)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{2x-4}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 2\\ &=\color{red}2 \end{aligned}\\\hline \color{blue}\textrm{Cara Kedua} \\\hline \begin{aligned}f(2)\: \,&=4\\ f(2+&h)=2(2+h)=4+2h\\ f(2+&h)-f(2)=2h\\ {f}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{2h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 2\\ &=\color{red}2 \end{aligned} \\\hline \end{array} \\ &\textrm{Jadi, laju perubahan fungsi}\: \: f\: \: \textrm{di}\\\ &x=2\: \: \textrm{adalah}\: \color{red}2 \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Jika} \: \: f(x)=3x-5, \textrm{hitunglah laju }\\ &\textrm{perubahan fungsi}\: \: f\: \: \textrm{di}\: \: x=2\\ &\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: \color{blue}f(x)=3x-5\\ &\begin{array}{|l|}\hline \color{blue}\textrm{Cara Pertama}\\\hline\begin{aligned} f(x)&=3x-5\\ f(2)&=3.2-5=1\\ {f}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=\color{red}3 \end{aligned}\\\hline \color{blue}\textrm{Cara Kedua} \\\hline \begin{aligned}f(2)\: \,&=1\\ f(2+&h)=3(2+h)-5=3h+1\\ f(2+&h)-f(2)=3h\\ {f}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(2+h)-f(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=\color{red}3 \end{aligned} \\\hline \end{array} \\ &\textrm{Jadi, laju perubahan fungsi}\: \: f\: \: \textrm{di}\\\ &x=2\: \: \textrm{adalah}\: \color{red}3 \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: f(x)=2022x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \color{purple}\begin{aligned}f(x)&=2022x^{2}\\ f(x+h)&=2022(x+h)^{2}\\ &=2022\left ( x^{2}+2xh+h^{2} \right )\\ &=2022x^{2}+4044xh+2022h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2022x^{2}+4044xh+2022h^{2} \right )-\left (2022x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4044xh+2022h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4044x+2022h\\ &=4044x \end{aligned}&\begin{aligned}{f}'(x)&=4044x\\ \textrm{maka},&\\ {f}'(1)&=4044.1\\ &=\color{red}4044\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa fungsi}\: \: f(x)=\displaystyle \frac{1}{x^{2}}\: \: \textrm{dengan daerah asal}\\ &\textrm{D}_{f}=\left \{ x\: |\: x\in \mathbb{R},\: \: \textrm{dan}\: \: x\neq 0 \right \}.\\ &\textrm{a}.\quad \textrm{Tunjukkan bahwa}\: \: {f}'(a)=\displaystyle -\frac{2}{a^{3}}\\ &\textrm{b}.\quad \textrm{jelaskanlah mengapa}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|l|l|}\hline \begin{aligned}{f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{(x+h)^{2}}-\displaystyle \frac{1}{x^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}-(x+h)^{2}}{\left (x(x+h) \right )^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-\left ( x^{2}+2xh+h^{2} \right )}{h\left ( x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2xh-h^{2}}{h\left (x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: -\displaystyle \frac{2x+h}{\left (x(x+h) \right )^{2}}\\ &=-\displaystyle \frac{2x}{x^{4}}\\ &=-\displaystyle \frac{2}{x^{3}} \end{aligned}&\begin{aligned}&\textrm{Untuk}\: \textrm{jawaban poin a dan b }\\ &\textrm{adalah sebagai berikut}\\ &{f}'(x)=-\displaystyle \frac{2}{x^{3}}\\ &{f}'(a)=-\displaystyle \frac{2}{a^{3}}\\ &\textrm{maka},\\ &{f}'(0)=-\displaystyle \frac{2}{0^{3}}\\ &=\color{red}-\displaystyle \frac{2}{0}\\ &\\ &\textrm{karena penyebut berupa }\\ &\textrm{bilangan}\: \: \color{red}0\\ &\textrm{maka}\: \: \color{red}{f}'(0)\: \: \color{black}\textbf{tidak terdefinisi}\\ &\\ &\\ & \end{aligned} \\\hline \end{array} \end{array}$.

Turunan Fungsi Aljabar

A. Turunan Fungsi Aljabar

A. 1 Laju Perubahasan untuk Nilai Fungsi

Konsep turunan fungsi pada awalnya digunakan dalam bidang kususnya Matematika dan fisika, dalam hal hal ini kita berikan contohnya adalah laju perubahan kecepatan.

Coba perhatikanlah, misal pada kasus gerak jatuh bebas suatu benda yang dinyaatakan dengan $\color{red}h=\color{black}\displaystyle \frac{1}{2}gt^{2}$ dengan $\color{red}h$ adalah tinggi benda dengan percepatan grafitasinya adalah $\color{red}g=\color{black}10\: \: m/s^{2}$ dan $\color{red}t$ adalah waktu tempuh.

Misalkan suatu benda jatuh dari ketinggian 125 meter dari permukaan tanah dengan percepatan grafitasinya adalah $g=10\: \: m/s^{2}$, maka waktu yang dibutuhkan benda tersebut untuk sampai ke tanah adalah:

$\begin{aligned}h&=\displaystyle \frac{1}{2}gt^{2}\\ 125&= \frac{1}{2}(10)t^{2}\\ 25&=t^{2}\\ 5&=t \end{aligned}$

Dari kejadian di atas dapat kita dapatkan kecepatan rata-ratanya yaitu: perubahan tinggi dibagi perubahan waktu terjadinya, atau misal dituliskan

$\bigtriangleup v=\displaystyle \frac{\bigtriangleup y}{\bigtriangleup t}=\displaystyle \frac{y_{n}-y_{1}}{t_{n}-t_{1}}$

Sehingga kecepatan rata-ratanya adalah : $\color{purple}\displaystyle \frac{125}{5}=25\: \: m/s^{2}$

Misalkan $\color{blue}f(t)$ untuk fungsi yang menujukkan posisi benda yang terjatuh dalam $\color{blue}t$ dengan $\color{blue}f(t)=5t^{2}$, maka kecepatan rata-ratanya kita dapat menghitungnya untuk beberapa selang termasuk kita dapat menghitung kecepatan sesaatnya.

Coba perhatikanlah tabel berikut:

$\begin{array}{|l|l|}\hline \begin{cases} f(4) &=5.4^{2}=80 \\ f(3) &= 5.3^{2}=45 \end{cases}&\color{purple}\begin{aligned}\bigtriangleup v&=\displaystyle \frac{80-45}{4-3}\\ &=\displaystyle \frac{35}{1}=35 \end{aligned}\\\hline \begin{cases} f(3,5) &=5.(3,5)^{2}=61,25 \\ f(3) &= 5.3^{2}=45 \end{cases}&\color{purple}\begin{aligned}\bigtriangleup v&=\displaystyle \frac{61,25-45}{3,5-3}\\ &=\displaystyle \frac{16,25}{0,5}=32,5 \end{aligned}\\\hline \begin{cases} f(3,25) &=5.(3,25)^{2}= \\ f(3) &= 5.3^{2}=45 \end{cases}&\color{purple}\begin{aligned}\bigtriangleup v&=\displaystyle \frac{52,8125-45}{3,25-3}\\ &=\displaystyle \frac{7,8125}{0,25}=31,25 \end{aligned}\\\hline \begin{cases} f(3,1) &=5.(3,1)^{2}=48,05 \\ f(3) &= 5.3^{2}=45 \end{cases}&\color{purple}\begin{aligned}\bigtriangleup v&=\displaystyle \frac{48,05-45}{3,1-3}\\ &=\displaystyle \frac{3,05}{0,1}=30,5 \end{aligned}\\\hline \begin{cases} f(3,1) &=5.(3,01)^{2}=45,3005 \\ f(3) &= 5.3^{2}=45 \end{cases}&\color{purple}\begin{aligned}\bigtriangleup v&=\displaystyle \frac{45,3005-45}{3,01-3}\\ &=\displaystyle \frac{0,3005}{0,01}=30,05 \end{aligned}\\\hline \end{array}$

Dari ilsutrasi tabel di atas jika selisih waktu diperkecil terus menerus sampai mendekati nol, maka kecepatan sesaatnya akan mendekati nilai 30.

Sehingga kecepatan ketika $t=3$ ditentukan sebagai laju perubahan jarak terhadap waktu yang dibutuhkan dapat dituliskan dengan:

$\begin{array}{|c|c|}\hline \textrm{Laju perubahan rata-rata}&\textrm{Laju perubahan sesaat}\\\hline \begin{aligned}&\\ &\displaystyle \frac{\Delta y}{\Delta x}=\displaystyle \frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}\\ & \end{aligned}&\begin{aligned}&\\ &\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(a+h)-f(a)}{h}\\ & \end{aligned}\\\hline \end{array}$.

Selanjutnya jika benda jatuh yang memenuhi kasus di atas, jika dihitung dengan pendekatan ini saat $t=3$ adalah:

$\color{red}\begin{aligned}\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(t+h)-f(t)}{h}&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{5(t+h)^{2}-5t^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{5\left ( t^{2}+2th+h^{2} \right )-5t^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{5t^{2}+10th+5h^{2}-5t^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{10th+5h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 10t+5h\\ &=10t \end{aligned}$

Dari saat $t=3$ kecepatan sesaatnya adalah $10t=10(3)=30\: \: m/s^{2}$.

Secara matematis, perubahan laju terhadap suatu fungsi di $x=a$ selanjutnya dinotasikan dengan $f'(x)$ dan didefiniskan dengan:

$\LARGE\color{purple}\boxed{f'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}}$

Bentuk di atas dinamakan dengan derivatif atau turunan pertama pada fungsi $f(x)$ dan dinotasikan dengan $f'(x)$ dan proses pencarian derivatif ini dinamakan differensial.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: g(x)=3x-5, \textrm{hitunglah laju perubahan }\\ &\textrm{fungsi}\: \: g\: \: \textrm{di}\: \: x=2\\ &\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\: \: \color{blue}g(x)=3x-5\\ &\begin{array}{|c|c|}\hline \color{purple}\textrm{Cara Pertama}&\color{purple}\textrm{Cara Kedua}\\\hline \begin{aligned} g(x)&=3x-5\\ g(2)&=3.2-5=1\\ {g}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{g(x)-g(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=\color{red}3 \end{aligned}&\begin{aligned}g(2)\: \,&=1\\ g(2+&h)=3(2+h)-5=3h+1\\ g(2+&h)-g(2)=3h\\ {g}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{g(2+h)-g(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=\color{red}3\\ & \end{aligned}\\\hline \end{array} \\ &\textrm{Jadi, laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\: \: \textrm{adalah}\: \color{red}3 \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: f(x)=2022x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \color{purple}\begin{aligned}f(x)&=2022x^{2}\\ f(x+h)&=2022(x+h)^{2}\\ &=2022\left ( x^{2}+2xh+h^{2} \right )\\ &=2022x^{2}+4044xh+2022h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2022x^{2}+4044xh+2022h^{2} \right )-\left (2022x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4044xh+2022h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4044x+2022h\\ &=4044x \end{aligned}&\begin{aligned}{f}'(x)&=4044x\\ \textrm{maka},&\\ {f}'(1)&=4044.1\\ &=\color{red}4044\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah kecepatan jika diketahui}\: \: f(t)=\sin t\\ &\textrm{saat}\: \: t\\\\ &\textrm{Jawab}:\\ &\color{purple}\begin{aligned}f'(t)=v(t)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(t+h)-f(t)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (t+h)-\sin t}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2\cos \displaystyle \frac{1}{2}(2t+h)\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 2\cos \displaystyle \frac{1}{2}(2t+h).\displaystyle \frac{\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 2\cos \displaystyle \frac{1}{2}(2t+h)\times \displaystyle \frac{1}{2}\\ &=2\cos \displaystyle \frac{1}{2}(2t+0)\times \displaystyle \frac{1}{2}\\ &=\cos \displaystyle \frac{1}{2}(2t)\\ &=\cos t \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui sebuah bola bergerak melingkar beraturan}\\ &\textrm{dengan persamaan}\: \: f(t)=2\sin 2t.\: \textrm{Tentukanlah}\\ &\textrm{kecepatan bola saat}\: \: t=\displaystyle \frac{1}{12}\pi \\\\ &\textrm{Jawab}:\\ &\color{purple}\begin{aligned}v(t)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(t+h)-f(t)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4\sin 2(t+h)-2\sin 2t}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4\cos \displaystyle \frac{1}{2}(4t+2h)\sin \displaystyle \frac{1}{2}(2h)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 4\cos \displaystyle \frac{1}{2}(4t+2h)\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h}\\ &=4\cos \displaystyle \frac{1}{2}(4t)\\ &=4\cos 2t\\ v\left ( \displaystyle \frac{1}{12}\pi \right )&=4\cos 2\left ( \displaystyle \frac{1}{12}\pi \right )\\ &=4\cos \displaystyle \frac{1}{6}\pi\\ &=4\left ( \displaystyle \frac{1}{2}\sqrt{3} \right ) \\ &=2\sqrt{3} \end{aligned} \end{array}$

DAFTAR PUSTAKA

Noormandiri, B. K. 2004. Matematika SMA Jilid $2^{A}$ Berdasarkan Kurikulum 2004. Jakarta: ERLANGGA.
Noormandiri, B. K. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 4 Soal dan Pembahasan Materi Lingkaran dan Hubungan Dua Lingkaran

$\begin{array}{ll}\\ 16.&\textrm{Salah satu garis singgung yang bersudut}\: \: 120^{\circ}\\ &\textrm{terhadap sumbu x positif terhadap lingkaran}\\ &\textrm{dengan ujung diameter titik}\: \: (7,6)\: \textrm{dan}\: \: (1,-2)\\ &\textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}y=-x\sqrt{3}+4\sqrt{3}+12\\ &\textrm{b}.\quad y=-x\sqrt{3}-4\sqrt{3}+8\\ &\textrm{c}.\quad y=-x\sqrt{3}+4\sqrt{3}-4\\ &\textrm{d}.\quad y=-x\sqrt{3}-4\sqrt{3}-8\\ &\textrm{e}.\quad y=-x\sqrt{3}+4\sqrt{3}+22\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline \textrm{Pusat Lingkaran}&\textrm{Gradien Garis Singgung}\\\hline \begin{aligned}&(a,b)\\ &=\left ( \displaystyle \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )\\ &=\left ( \displaystyle \frac{7+1}{2},\frac{6+(-2)}{2} \right )\\ &=(4,2) \end{aligned}&\begin{aligned}m&=\tan 120^{\circ}\\ &=-\tan \left ( 180^{\circ}-60^{\circ} \right )\\ &=-\tan 60^{\circ}\\ &=-\sqrt{3}\\ &\\ \end{aligned} \\\hline \textrm{Jari-jari}&\textrm{Garis Singgung}\\\hline \begin{aligned}r&=\textrm{jarak titik}\\ &\: \: \: \: \: \, \textrm{singgung ke pusat}\\ &=\sqrt{(7-4)^{2}+(6-2)^{2}}\\ &=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{25}\\ &=5\\ &\\ &\\ & \end{aligned}&\begin{aligned} &(y-b)=m(x-a)\pm r\sqrt{1+m^{2}}\\ &\Leftrightarrow (y-2)=-\sqrt{3}(x-4)\pm 5\sqrt{1+(-\sqrt{3})^{2}}\\ &\Leftrightarrow y-2=-\sqrt{3}x+4\sqrt{3}\pm 5\sqrt{1+4}\\ &\Leftrightarrow y=-\sqrt{3}x+4\sqrt{3}+2\pm 10\\ &\Leftrightarrow y=\begin{cases} -\sqrt{3}x+4\sqrt{3}+2+ 10 \\ -\sqrt{3}x+4\sqrt{3}+2- 10 \end{cases}\\ &\Leftrightarrow y=\begin{cases} \color{red}-\sqrt{3}x+4\sqrt{3}+12 & \\ -\sqrt{3}x+4\sqrt{3}-8 & \end{cases} \end{aligned}\\\hline \end{array}\\ &\textrm{Berikut ilustrasi gambarnya} \end{array}$.

Dengan ilustrasi tambahan

$\begin{array}{ll}\\ 17.&\textrm{Salah satu garis singgung lingkaran}\\\ & x^{2}+y^{2}=10\: \: \textrm{yang ditarik dari}\\ &\textrm{titik}\: \: (4,2)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \color{red}x+3y=10\\ &\textrm{b}.\quad x-3y=10\\ &\textrm{c}.\quad -x-3y=10\\ &\textrm{d}.\quad 2x+y=10\\ &\textrm{e}.\quad x+2y=10\\\\ &\textbf{Jawab}:\\ &\begin{array}{|c|c|}\hline \begin{aligned}&\textrm{Garis Singgung}\\ &\quad\quad \textrm{di titik}\\ &(x_{1},y_{1})=(4,2) \end{aligned}&\begin{aligned}&\textrm{Tahapan menentukan}\\ &\quad\qquad \textrm{harga}\: \: m\\ & \end{aligned}\\\hline \begin{aligned}&y-y_{1}=m(x-x_{1})\\ &y-2=m(x-4)\\ &y=mx-4m+2\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&x^{2}+y^{2}=10\\ &x^{2}+\left ( mx-4m+2 \right )^{2}=10\\ &x^{2}+m^{2}x^{2}+16m^{2}+4-8m^{2}x+4mx-16m=10\\ &x^{2}+m^{2}x^{2}+16m^{2}-8m^{2}x+4mx-16m-6=0\\ &(1+m^{2})x^{2}+(4m-8m^{2})x+16m^{2}-16m-6=0\\ &\begin{cases} a & =1+m^{2} \\ b & =4m-8m^{2} \\ c & =16m^{2}-16m-6 \end{cases} \end{aligned}\\\hline \end{array}\\ &\begin{aligned}&\textrm{Syarat menyinggung}\: \: D=0\\ &b^{2}-4ac=0\\ &\left ( 4m-8m^{2} \right )^{2}-4\left ( 1+m^{2} \right )\left ( 16m^{2}-16m-6 \right )=0\\ &16m^{2}-64m^{3}+64m^{4}-64m^{2}+64m+24-64m^{4}+64m^{3}+24m^{2}=0\\ &-24m^{2}+64m+24=0\\ &-3m^{2}+8m+3=0\\ &(m-3)(3m+1)=0\\ &m=3\: \: \textrm{atau}\: \: m=-\displaystyle \frac{1}{3}\\ &m=\begin{cases} 3 & \Rightarrow y=3x-10\\ &\Rightarrow 3x-y=10\\ -\displaystyle \frac{1}{3} & \Rightarrow y=-\displaystyle \frac{1}{3}x+\frac{4}{3}+2\\ &\Rightarrow \color{red}x+3y=10 \end{cases} \end{aligned} \end{array}$.

$.\qquad\begin{aligned}&\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned}$

$\begin{array}{ll}\\ 18.&\textrm{Diketahui persamaan lingkaran}\: \: x^{2}+y^{2}=r^{2}\\ &\textrm{dan sebuah titik di luar lingkaran}\: \: M(a,b)\\ &\textrm{Posisi garis}\: \: ax+by=r^{2}\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{menyinggung lingkaran}\\ &\textrm{b}.\quad \color{red}\textrm{memotong lingkaran di dua titik}\\ &\textrm{c}.\quad \textrm{melalui titik pusat lingkaran}\\ &\textrm{d}.\quad \textrm{tidak memotong lingkaran}\\ &\textrm{e}.\quad \textrm{tidak ada yang benar}\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa}\\ &\bullet \quad L\equiv x^{2}+y^{2}=r^{2}\\ &\bullet \quad M(a,b)\: \: \textrm{di luar lingkaran}\: \: L\\ &\color{purple}\textrm{Selanjutnya perhatikan penjelasan berikut}\\ &\begin{aligned}&\textrm{Karena}\: M(a,b)\: \textrm{di luar lingkaran}\: L,\: \textrm{maka}\\ &\textrm{maka salah satu dari}\: \: a\: \: \textrm{atau}\: \: b\: \: \textrm{atau keduanya}\\ &\textrm{akan lebih besar nilanya dari pada}\: \: r.\\ &\textrm{Misalkan kita pilih}\: \: a>r\\ &\color{blue}\textrm{Ambil posisi saat memotong sumbu}-X,\: \color{black}y=0\\ &\begin{aligned}&\textrm{Untuk lingkaran}\: \: x^{2}+y^{2}=r^{2}\\ &\bullet \quad y=0\Rightarrow x^{2}+0^{2}=r^{2}\Rightarrow x=\left | r \right |\\ &\textrm{Untuk garis}\: \: ax+by=r^{2}\\ &\bullet \quad y=0\Rightarrow ax=r^{2}\Rightarrow x=\displaystyle \frac{r^{2}}{a}\\ &\textrm{Dari sini tampak posisi}\: \: x=\color{red}\left | r \right |> \displaystyle \frac{r^{2}}{a}\geq 0 \end{aligned}\\ &\textrm{Sehingga kesimpulannya adalah}:\\ &\color{red}\textrm{garis tersebut akan selalu memotong lingkaran} \end{aligned}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Dua lingkaran dengan persamaan}\\ &\textrm{lingkaran-lingkaran}\: x^{2}+y^{2}+6x-8y+21=0\\ &\textrm{dan}\: \: x^{2}+y^{2}+10x-8y+25=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{berpotongan di luar titik}\\ &\textrm{b}.\quad \textrm{tidak berpotongan atau bersinggungan}\\ &\textrm{c}.\quad \textrm{bersinggungan luar}\\ &\textrm{d}.\quad \color{red}\textrm{bersinggungan dalam}\\ &\textrm{e}.\quad \textrm{sepusat}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+6x-8y+21=0&\begin{cases} P_{1} &=(-3,4) \\ r_{1} & = 2 \end{cases}\\\hline L_{2}\equiv x^{2}+y^{2}+10x-8y+25=0&\begin{cases} P_{2} &=(-5,4) \\ r_{2} & = 4 \end{cases}\\\hline \end{array} \\ &\textrm{dan}\\ &\begin{array}{|c|c|}\hline \textrm{Jarak kedua pusat}&\textrm{Jumlah/selisih jari-jari}\\\hline \begin{aligned}&\left (P_{1}P_{2} \right )\\ &=\sqrt{(-3+5)^{2}+(4-4)^{2}}\\ &=\sqrt{2^{2}+0^{2}}=\sqrt{4}=2 \end{aligned}&\begin{aligned}\begin{cases} r_{1}+r_{2} & =2+4=6 \\ \left |r_{1}-r_{2} \right | & =\left | 2-4 \right |=2 \end{cases} \end{aligned}\\\hline \end{array}\\ &\textrm{Karena nilai}\: \: \color{red}P_{1}P_{2}\color{black}=\color{red}\left |r_{1}-r_{2} \right |\color{black}=\color{red}2\\ &\textrm{hal ini menunjukkan keduanya bersinggungan}\\ &\color{blue}\textrm{di dalam}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$ .

$\begin{array}{ll}\\ 20.&\textrm{Dua lingkaran dengan persamaan}\\ &\textrm{lingkaran-lingkaran}\: x^{2}+y^{2}+2x-6y+9=0\\ &\textrm{dan}\: \: x^{2}+y^{2}+8x-6y+9=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \textrm{berpotongan}\\ &\textrm{b}.\quad \color{red}\textrm{bersinggungan di dalam}\\ &\textrm{c}.\quad \textrm{bersinggungan luar}\\ &\textrm{d}.\quad \textrm{tidak berpotongan}\\ &\textrm{e}.\quad \textrm{sepusat}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan bahwa}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+2x-6y+9=0&\begin{cases} P_{1} &=(-1,3) \\ r_{1} & = 1 \end{cases}\\\hline L_{2}\equiv x^{2}+y^{2}+8x-6y+9=0&\begin{cases} P_{2} &=(-4,3) \\ r_{2} & = 4 \end{cases}\\\hline \end{array} \\ &\textrm{dan}\\ &\begin{array}{|c|c|}\hline \textrm{Jarak kedua pusat}&\textrm{Jumlah/selisih jari-jari}\\\hline \begin{aligned}&\left (P_{1}P_{2} \right )\\ &=\sqrt{(-1+4)^{2}+(3-3)^{2}}\\ &=\sqrt{3^{2}+0^{2}}=\sqrt{9}=3 \end{aligned}&\begin{aligned}\begin{cases} r_{1}+r_{2} & =1+4=5 \\ \left |r_{1}-r_{2} \right | & =\left | 1-4 \right |=3 \end{cases} \end{aligned}\\\hline \end{array}\\ &\textrm{Karena nilai}\: \: \color{red}P_{1}P_{2}\color{black}=\color{red}\left |r_{1}-r_{2} \right |\color{black}=\color{red}3\\ &\textrm{hal ini menunjukkan keduanya bersinggungan}\\ &\color{blue}\textrm{di dalam}\\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$.

DAFTAR PUSTAKA

Budi, W. S. 2010. Bahan Ajar Persiapan Menuju Olimpiade Sain Nasional/Internasional Matematika 3. Jakarta: ZAMRUD KEMALA.
Kartini, Suprapto, Subandi, dan Setiadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
Kanginan M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
Noormandiri. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU
Sukino. 2017. Matematika Jilid 2 untuk Kelas SMA/MA Kelas XI Kelompok Peminatan dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Contoh 3 Soal dan Pembahasan Materi Lingkaran

$\begin{array}{ll}\\ 11.&\textrm{Lingkaran}\: \: x^{2}+y^{2}+2ax+2by+c=0\\ &\textrm{menyinggung sumbu Y jika}\: \: c\: =....\\ &\textrm{A}.\quad ab\\ &\textrm{B}.\quad ab^{2}\\ &\textrm{C}.\quad a^{2}b\\ &\textrm{D}.\quad a^{2}\\ &\textrm{E}.\quad \color{red}b^{2}\\\\ &\textbf{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\begin{aligned}&x^{2}+y^{2}+2ax+2by+c=0\\ &x=0\Rightarrow 0^{2}+y^{2}+2a.0+2by+c=0\\ &y^{2}+2by+c=0\begin{cases} a & =1 \\ b & =2b \\ c & =c \end{cases}\\ &\textrm{Syarat menyinggung}\: \textrm{adalah}:\\ &D=b^{2}-4ac=0\\ &\Leftrightarrow (2b)^{2}-4.1.c=0\\ &\Leftrightarrow 4c=4b^{2}\\ &\Leftrightarrow c=\color{red}b^{2} \end{aligned} \\\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&x^{2}+y^{2}+2ax+2by+c=0\\ &\Leftrightarrow x^{2}+2ax+a^{2}+y^{2}+2by+b^{2}+c-a^{2}-b^{2}=0\\ &\Leftrightarrow (x+a)^{2}+(y+b)^{2}=a^{2}+b^{2}-c\\ &\textrm{Karena menyinggung sumbu-Y, maka}\: \: R=a \\ &\textrm{Sehingga}\: \: R^{2}=a^{2}+b^{2}-c=a^{2}\\ &\Leftrightarrow b^{2}-c=0\\ &\Leftrightarrow b^{2}=c\\ &\Leftrightarrow c=\color{red}b^{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Diketahui pusat lingkaran L terletak dikuadran}\\ &\textrm{I dan berada di sepanjang garis}\: \: y=2x.\: \: \textrm{Jika}\\ &\textrm{lingkaran L menyinggung sumbu Y di titik}\\ &(0,6),\: \textrm{maka persamaan lingkaran L adalah}\: ....\\ &\textrm{A}.\quad x^{2}+y^{2}-3x-6y=0\\ &\textrm{B}.\quad x^{2}+y^{2}+6x+12y-108=0\\ &\textrm{C}.\quad x^{2}+y^{2}+12x+6y-72=0\\ &\textrm{D}.\quad x^{2}+y^{2}-12x-6y=0\\ &\textrm{E}.\quad \color{red}x^{2}+y^{2}-6x-12y+36=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&(x-a)^{2}+(y-b)^{2}=r^{2},\\ &\textrm{menyinggung titik}\: \: (0,6)\\ &\textrm{berarti pusat lingkaran L juga terletak}\\ &\textrm{pada garis}\: \: y=6.\: \: \textrm{Hal ini menunjukkan bahwa }\\ &\textrm{pusat lingkaran}\: \: \, \: \textrm{L berpusat di}\: \: (x,2x)=(\frac{y}{2},y),\\ &\textrm{dengan}\: \: y=6.\, \: \textrm{Dari informasi di atas, }\\ &\textrm{didapatlah pusat lingkaran berada di titik}\: \: (3,6).\\ &\textrm{Sehingga persamaan lingkarannya adalah}:\\ &(x-3)^{2}+(y-6)^{2}=3^{2}\: \: \textrm{ingat}\: \: r=\textrm{absis}\: \: x=3\\ &\Leftrightarrow (x-3)^{2}+(y-6)^{2}=x^{2}-6x+9+y^{2}+12x+36=9\\ &\Leftrightarrow \, \color{red}x^{2}+y^{2}-6x+12y+36=0\\ &\color{purle}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Persamaan garis singgung lingkaran}\\ &x^{2}+y^{2}+8x-3y-24=0,\: \: \textrm{di titik}\\ & (2,4)\: \: \textrm{adalah}\: ....\\ &\textrm{A}.\quad 12x-5y-44=0\\ &\textrm{B}.\quad \color{red}12x+5y-44=0\\ &\textrm{C}.\quad 12x-y-50=0\\ &\textrm{D}.\quad 12x+y-50=0\\ &\textrm{E}.\quad 12x+y+50=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&x^{2}+y^{2}+8x-3y-24\\ &\Leftrightarrow x^{2}+8x+16+y^{2}-3y+\displaystyle \frac{9}{4}-24=16+\frac{9}{4}\\ &\Leftrightarrow \: (x+4)^{2}+(y-\frac{3}{2})^{2}=16+\frac{9}{4}+24=42\frac{1}{4}\\ &\textrm{Persamaan garis singgung lingkar}\textrm{an lingkaran }\\ &\textrm{di titik}\: \: (x_{1},y_{1})\: \: \textrm{adalah}:\\ &(x_{1}+4)(x+4)+(y_{1}-\frac{3}{2})(y-\frac{3}{2})=42\frac{1}{4},\\ &\textrm{untuk}\: \: (x_{1},y_{1})=(2,4),\: \textrm{maka}\\ &(2+4)(x+4)+(4-\frac{3}{2})(y-\frac{3}{2})=\frac{169}{4}\\ &\Leftrightarrow 6(x+4)+\frac{5}{2}(y-\frac{3}{2})=\frac{169}{4}\\ &\Leftrightarrow 24(x+4)+5(2y-3)=169\\ &\Leftrightarrow 24x+96+10y-15=169\\ &\Leftrightarrow 24x+10y=169-96+15=88\\ &\Leftrightarrow \color{red}12x+5y-44=0\\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Sebuah garis singgung}\: \: g\: \: \textrm{menyinggung }\\ &\textrm{lingkaran yang berpusat di}\: \: (-2,5)\: \: \textrm{dan}\\ &\textrm{berjari-jari}\: \: 2\sqrt{10}\: \: \textrm{di titk}\: \: (4,3),\: \textrm{maka }\\ &\textrm{persamaan garis singgung}\: \: g\: \: \textrm{adalah}\: .... \\ &\textrm{A}.\quad y=3x+9\\ &\textrm{B}.\quad \color{red}y=3x-9\\ &\textrm{C}.\quad y=-3x+9\\ &\textrm{D}.\quad y=-3x-9\\ &\textrm{E}.\quad y=3x+21\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&(x-a)^{2}+(y-b)^{2}=r^{2}\\ &\begin{cases} \textrm{Pusat} & =(-2,5) \\ \textrm{r} & =2\sqrt{10} \end{cases} \\ &\textrm{maka persamaan lingkarannya}:\\ &(x+2)^{2}+(y-5)^{2}=(2\sqrt{10})^{2}\\ &\Leftrightarrow (x_{1}+2)(x+2)+(y_{1}-5)(y-5)=40,\\ &\textrm{menyingung garis}\: \: g\: \: \textrm{di}\: (4,3)\\ &(4+2)(x+2)+(3-5)(y-5)=40\\ &\Leftrightarrow 6x+12-2y+10=40\\ &\Leftrightarrow 6x-2y=40-12-10\\ &\Leftrightarrow 3x-y=9\\ &\Leftrightarrow -y=-3x+9\\ &\Leftrightarrow \color{red}y=3x-9\\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Suatu lingkaran dengan titik pusatnya terletak }\\ &\textrm{pada kurva}\: \: y=\sqrt{x}\: \: \textrm{dan melalui titik asal}\: \: O(0,0).\\ & \textrm{Jika diketahui absis titik pusat lingkaran tersebut }\\ &\textrm{adalah}\: \: a,\: \: \textrm{maka persamaan garis singgung }\\ &\textrm{lingkaran yang melalui titik}\: \: O\: \: \textrm{tersebut adalah}\: ....\\ &\textrm{A}.\quad y=-x\\ &\textrm{B}.\quad \color{red}y=-x\sqrt{a}\\ &\textrm{C}.\quad y=-ax\\ &\textrm{D}.\quad y=-2x\sqrt{2}\\ &\textrm{E}.\quad y=-2ax\\\\ &\textbf{Jawab}:\\ &\begin{array}{|l|c|l|}\hline \begin{aligned}&\textrm{Pusat}\\ &\textrm{lingkaran}\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Gradien garis singgung}\\ &\textrm{yang tegak lurus dengan }\\ &\textrm{garis yang melalui titik}\\ &\textrm{pusat lingkaran yang }\\ &\textrm{bergradien}\: \: m_{L} \end{aligned}&\begin{aligned}&\textrm{Persamaan garis }\\ &\textrm{singgung yang }\\ &\textrm{melalui titik asal}\\ &O(0,0)\\ & \end{aligned}\\\hline \begin{aligned}&(a,b)\\ &=\left ( a,\sqrt{a} \right )\\ &\\ &\\ & \end{aligned}&\color{blue}\begin{aligned}&m.m_{1}=-1\\ &m.\frac{y}{x}=-1\\ &m=-\frac{x}{y}=-\displaystyle \frac{a}{\sqrt{a}}\\ &\: \: \: \, =-\sqrt{a} \end{aligned}&\begin{aligned}y&=mx,\\ & \textrm{karena melalui}\\ &\textrm{titik asal}\\ y&=-\sqrt{a}x,\\ y&=\color{red}-x\sqrt{a} \end{aligned}\\\hline \end{array} \end{array}$.