Contoh Soal 4 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 16.& \text{Jika grafik fungsi}f(x)=5+15x+9x^2+x^3\\ & \text{naik untuk}x\text{yang memenuhi}....\\ & \begin{array}{l}\\ \text{a}.x<1\text{atau}x>5& \text{d}.x<-5\text{atau}x>-1\\ \text{b}.-{\displaystyle 1<x<5}& \text{e}.-5<x<1\\ \text{c}.{\displaystyle -5<x<-1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=x^3+9x^2+15x+5.\\ & \text{Dikatakan fungsi}f\text{naik, maka}\\ & f{}^{\prime }(x)>0\\ & \iff 3x^2+18x+15>0,\text{tiap ruas dibagi 3}\\ & \iff x^2+6x+5>0\\ & \iff (x+1)(x+5)>0\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Sebuah bola dilempar ke atas secara vertikal.}\\ & \text{Jika lintasan bola pada saat}t\text{detik adalah}\\ & h(t)=-{\displaystyle \frac{1}{4}{t}^4+\frac{2}{3}{t}^3+4t^2+5\text{m},}\\ & \text{maka tinggi maksimum yang}\\ & \text{dicapai oleh bola tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{67}{3}}& & \text{d}.{\displaystyle \frac{133}{3}}\\ \\ \text{b}.{\displaystyle \frac{123}{3}}& \text{c}.{\displaystyle \frac{128}{3}}& \text{e}.{\displaystyle \frac{143}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perh}& \text{atikan bahwa lintasan bola saat dilempar }\\ \text{verti}& \text{kal dituliskan dengan fungsi}\\ h(t)& =-{\displaystyle \frac{1}{4}{t}^4+\frac{2}{3}{t}^3+4t^2+5.}\\ \text{mak}& \text{a tinggi maksimum akan dicapai bola }\\ \text{saat}& h{}^{\prime }(t)=0,\\ \text{Sela}& \text{njutnya},\\ h{}^{\prime }(t)& =0\\ -t^3+& 2t^2+8t=0\\ -t(t^2& -2t-8)=0\\ t(t-& 4)(t+2)=0\{\begin{array}{ll}t& =0\\ t& =4\\ t& =-2\end{array}\\ \text{kita}& \text{ambil yang bernilai positif untuk }\\ & t\text{yaitu}t=4.\\ t=4& \to h(4)=-{\displaystyle \frac{1}{4}(4{)}^4+\frac{2}{3}(4{)}^3+4(4{)}^2+5}\\ h(4)& =-64+{\displaystyle \frac{128}{3}+64+5}\\ & ={\displaystyle \frac{143}{3}m}\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

Contoh Soal 3 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 11.& \text{Jika}f(x)={\displaystyle (x^2+2)\sqrt{x^2+x+3}}\\ & \text{maka}f{}^{\prime }(2)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18}& & \text{d}.13\\ \\ \text{b}.{\displaystyle 17}& \text{c}.{\displaystyle 15}& \text{e}.12\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle (x^2+2)\sqrt{x^2+x+3}}\\ f{}^{\prime }(x)& =(2x)\sqrt{x^2+x+3}\\ & +(x^2+2).{\displaystyle \frac{1}{2}{(x^2+x+3)}^{{}^{-\frac{1}{2}}}.(2x+1)}\\ & =2x\sqrt{x^2+x+3}+{\displaystyle \frac{(x^2+2)(2x+1)}{2\sqrt{x^2+x+3}}}\\ f{}^{\prime }(2)& =2(2)\sqrt{(2{)}^2+(2)+3}+{\displaystyle \frac{((2{)}^2+2)(2(2)+1)}{2\sqrt{(2{)}^2+(2)+3}}}\\ & =4\sqrt{9}+{\displaystyle \frac{6.5}{2\sqrt{9}}}\\ & =4.3+{\displaystyle \frac{6.5}{2.3}}\\ & =12+5\\ & =17\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Jika rusuk suatu kubus bertambah panjang }\\ & \text{dengan laju 7}cm/detik,\text{maka laju}\\ & \text{bertambahnya volume pada saat rusuk }\\ & \text{panjangnya 15}cm\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 675c{m}^3/detik}& \text{d}.4725c{m}^3/detik\\ \text{b}.{\displaystyle 1575c{m}^3/detik}& \text{e}.23625c{m}^3/detik\\ \text{c}.{\displaystyle 3375c{m}^3/detik}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Laju }& \text{pertambahan volumenya}:\\ & \{\begin{array}{ll}\bullet \frac{\mathrm{d}s}{\mathrm{d}t}& =7cm/detik\\ \bullet V& =s^3\to \text{d}V=3s^2\text{d}s\text{atau}\\ \frac{\mathrm{d}V}{\mathrm{d}s}& =3s^{2}c{m}^3/cm\to s=15cm\end{array}\\ \frac{\mathrm{d}V}{\mathrm{d}t}& =\frac{\mathrm{d}V}{\mathrm{d}t}\\ & =\frac{\mathrm{d}V}{\mathrm{d}s}\times \frac{\mathrm{d}s}{\mathrm{d}t}\\ & =3s^{2}c{m}^3/cm\times 7cm/detik\\ & =3(15{)}^2\times 7c{m}^3/detik\\ & =4725c{m}^3/detik\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{persamaan garis singgung di}x=1\\ & \text{pada kurva}y=x^3-3x^2+1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.y=-3x+2& \text{d}.y=3x-2\\ \text{b}.y=-3x+4& \text{e}.y=-3x+3\\ \text{c}.y=3x-4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|c|}\hline \text{Titik singgung di}x=1\\ \begin{array}{rl}y& =x^3-3x^2+1\\ y& =(1{)}^3-3(1{)}^2+1\\ & =1-3+1\\ & =-1\\ & \text{titik}(a,b)=(1,-1)\end{array}\\ \text{Gradien garis singgung di}x=1\\ \begin{array}{rl}y{}^{\prime }=m{}_{{}_{x=1}}& =3x^2-6x\\ & =3(1{)}^2-6(1)\\ & =3-6\\ & =-3\end{array}\\ \text{Persamaan garis singgung}\\ \begin{array}{rl}y& =m(x-a)+b\\ & =-3(x-1)+(-1)\\ & =-3x+3-1\\ & =-3x+2\end{array}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Suatu kurva}y=x^3+2a{x}^2+b\\ & \text{Sebuah garis}y=-9x-2\text{menyinggung}\\ & \text{kurva di titik dengan}x=1,\\ & \text{maka nilai}a\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.-3& & \text{d}.3\\ \text{b}.-{\displaystyle \frac{1}{3}}& \text{c}.{\displaystyle \frac{1}{3}}& \text{e}.8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Gradien garis singgung}\\ & y{}_{{}_{{}_{dix=1}}}\{\begin{array}{ll}y& =x^3+2a{x}^2+b\\ & \to m=y{}^{\prime }=3x^2+4ax\\ \\ y& =-9x-2\to m=y^{\prime }=-9\end{array}\\ & \text{Sehingga},\\ & m=m=y{}^{\prime }\\ & -9=3x^2+4ax\\ & -9=3(1{)}^2+4a(1)\\ & -9=3+4a\\ & -4a=3+9\\ & a={\displaystyle \frac{12}{-4}}\\ & =-3\\ & \end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika grafik fungsi}f(x)=x^3+a{x}^2+bx+c\\ & \text{hanya turun untuk interval}-1<x<5,\\ & \text{maka nilai}a+b\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.-21& & \text{d}.21\\ \text{b}.-{\displaystyle 9}& \text{c}.{\displaystyle 9}& \text{e}.24\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=x^3+a{x}^2+bx+c\\ & \text{grafik fungsi turun}\text{berarti}:f{}^{\prime }(x)<0\\ & f{}^{\prime }(x)<0\\ & \iff 3x^2+2ax+b<0\\ & \iff (x+1)(x-5)<0\\ & \text{ini maksud pada interval}\\ & -1<x<5\text{pada soal}\\ & x^2-4x-5<0\\ & \text{dikalikan dengan 3 supaya terjadi persamaan}\\ & 3x^2-3.4x-3.5=3x^2+2ax+b<0\\ & 3x^2+2(-6)x+(-15)=3x^2+2ax+b<0\\ & \{\begin{array}{ll}a& =-6\\ b& =-15\end{array}\\ & \text{Sehingga},\\ & a+b=-6+(-15)=-21\end{array}\end{array}$.

Contoh Soal 2 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 6.& \text{Jika}W=\sin 2t,\text{maka}{\displaystyle \frac{dW}{dt}=....}\\ & \begin{array}{l}\\ \text{a}.\cos 2t& \text{d}.2t\cos 2t+\sin 2t\\ \text{b}.2\cos 2t& \text{e}.\sin 2t-t\cos 2t\\ \text{c}.\sin 2t+t\cos 2t\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & W=\sin 2t\\ & W=\sin u\text{dengan}u=2t\\ & {\displaystyle \frac{dW}{dt}={\displaystyle \frac{dW}{du}.\frac{du}{dt}}}\\ & =\cos u.2\\ & =2\cos u\\ & =2\cos 2t\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}f(x)={\displaystyle \frac{2x+4}{1+\sqrt{x}},\text{maka}{\displaystyle f{}^{\prime }(4)=....}}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}}& & \text{d}.1\\ \text{b}.{\displaystyle \frac{3}{7}}& \text{c}.{\displaystyle \frac{3}{5}}& \text{e}.4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{2x+4}{1+\sqrt{x}}}\\ & ={\displaystyle \frac{U}{V}}\\ f{}^{\prime }(x)& ={\displaystyle \frac{U{}^{\prime }.V-U.V{}^{\prime }}{V^2}}\\ & ={\displaystyle \frac{(2)(1+\sqrt{x})-(2x+4).(1.{(1+\sqrt{x})}^0.{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}})}{{(1+\sqrt{x})}^2}}\\ & \text{ingat}\sqrt{x}=x^{{}^{\frac{1}{2}}}\\ f{}^{\prime }(4)& ={\displaystyle \frac{2(1+\sqrt{4})-(2.4+4).\frac{1}{2}.4^{{}^{-\frac{1}{2}}}}{{(1+\sqrt{4})}^2}}\\ & ={\displaystyle \frac{2(1+2)-(12).\frac{1}{2}.\frac{1}{2}}{(1+2{)}^2}}\\ & \text{ingat juga}{4}^{{}^{-\frac{1}{2}}}={\left(2^2\right)}^{{}^{-\frac{1}{2}}}=2^{{}^{-1}}={\displaystyle \frac{1}{2^1}=\frac{1}{2}}\\ & ={\displaystyle \frac{6-3}{9}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Jika}f(x)={\displaystyle \frac{1+\sin x}{\cos x},\text{maka}f{}^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}}& & \text{d}.2\\ \text{b}.{\displaystyle -\frac{1}{2}}& \text{c}.{\displaystyle \frac{3}{4}}& \text{e}.-2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{1+\sin x}{\cos x}}\\ f{}^{\prime }(x)& ={\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & \text{ingat bahwa}{\sin }^{2}x+{\cos }^{2}x=1\\ & ={\displaystyle \frac{1+\sin x}{{\cos }^{2}x}}\\ f{}^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)& ={\displaystyle \frac{1+\sin \left({\displaystyle \frac{1}{6}\pi }\right)}{{\cos }^2\left({\displaystyle \frac{1}{6}\pi }\right)}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{2}}}{{\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}^2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{3}{4}}}}\\ & ={\displaystyle \frac{3}{2}\times \frac{4}{3}}\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}f(x)={\displaystyle 3x^2-2ax+7}\\ & \text{dan}f{}^{\prime }(1)=0,\text{maka}f{}^{\prime }(2)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}& & \text{d}.6\\ \text{b}.{\displaystyle 2}& \text{c}.{\displaystyle 4}& \text{e}.8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle 3x^2-2ax+7}\\ f{}^{\prime }(x)& =6x-2a\\ f{}^{\prime }(1)& =0\\ 6(1)-2a& =0\\ 6& =2a\\ 3& =a\\ \text{sehingga}& \\ f{}^{\prime }(x)& =6x-6\\ \text{maka},& \\ f{}^{\prime }(2)& =6.2-6\\ & =12-6\\ & =6\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika}f(x)={\displaystyle (6x-3{)}^3(2x-1)}\\ & \text{maka}f{}^{\prime }(1)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18}& & \text{d}.162\\ \\ \text{b}.{\displaystyle 24}& \text{c}.{\displaystyle 54}& \text{e}.216\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle (6x-3{)}^3(2x-1)}\\ & =(3.(2x-1){)}^3(2x-1{)}^1\\ & =3^3.(2x-1{)}^{3+1}\\ & =27(2x-1{)}^4\\ f{}^{\prime }(x)& =4.27(2x-1{)}^{4-1}.2\\ & =216.(2x-1{)}^3\\ f{}^{\prime }(1)& =216.(2.1-1{)}^3\\ & =216.1\\ & =216\end{array}\end{array}$.

Contoh Soal 1 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 1.& \text{Diketahui}f{}^{\prime }(2)=\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{x^3-8}{x-2},}\\ & \text{maka fungsi}f(x)=....\\ & \begin{array}{l}\\ \text{a}.12& & \text{d}.x^3\\ \text{b}.2& \text{c}.x& \text{e}.x-8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Turunan fungsi f di}x=c\text{adalah}\\ & f{}^{\prime }(c)=\underset{x\to c}{\text{Lim}}{\displaystyle \frac{f(x)-f(c)}{x-c},\text{maka}}\\ & \text{turunan fungsi f di}x=2\text{adalah}\\ & f{}^{\prime }(2)=\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{x^3-8}{x-2}}\\ & \text{sehingga akan didapa}\text{tkan fungsi}f\text{nya }\\ & \text{yaitu}f(x)=x^3\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}a\ne 0,\text{maka nilai dari}\\ & \underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}=....}\\ & \begin{array}{l}\\ \text{a}.3a{\displaystyle \sqrt[3]{a}}& & \text{d}.{\displaystyle \frac{1}{2a}\sqrt[3]{a}}\\ \text{b}.2a{\displaystyle \sqrt[3]{a}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{3a}\sqrt[3]{a}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}f{}^{\prime }(a)& =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{(\sqrt[3]{x}-\sqrt[3]{a})(\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2})}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{1}{(\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2})}}\\ & ={\displaystyle \frac{1}{(\sqrt[3]{a^2}+\sqrt[3]{a.a}+\sqrt[3]{a^2})}}\\ & ={\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{array}\\ & \text{Alternatif 2 (dengan aturan L'Hopital)}\\ & \begin{array}{rl}\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}}& =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{f(x)}{g(x)}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{f{}^{\prime }(x)}{g{}^{\prime }(x)}}\\ \underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}}& =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{x^{{}^{\frac{1}{3}}}-a^{{}^{\frac{1}{3}}}}{x-a}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{3}{x}^{{}^{\frac{1}{3}-1}}-0}}{1-0}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{3}{x}^{{}^{-\frac{2}{3}}}}}{1}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{1}{3\sqrt[3]{x^2}}\times \frac{\sqrt[3]{x}}{\sqrt[3]{x}}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{1}{3x}\sqrt[3]{x}}\\ & ={\displaystyle \frac{1}{3a}\sqrt[3]{a}}\end{array}\end{array}$ .

$\begin{array}{l}\\ 3.& \text{Jika}f(x)={\displaystyle \frac{1}{\sqrt{x}}\text{maka nilai dari}}\\ & -2f{}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{x\sqrt{x}}}& & \text{d}.{\displaystyle -\frac{1}{2x\sqrt{x}}}\\ \text{b}.{\displaystyle x\sqrt{x}}& \text{c}.{\displaystyle -\frac{1}{2\sqrt{x}}}& \text{e}.{\displaystyle -2x\sqrt{x}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dike}& \text{tahui}\\ f(x)& ={\displaystyle \frac{1}{\sqrt{x}}={\displaystyle \frac{1}{x^{{}^{\frac{1}{2}}}}=x^{{}^{-\frac{1}{2}}}}}\end{array}\\ & \begin{array}{|cc|}\hline y=a{x}^n\to y{}^{\prime }=na{x}^{n-1}& \begin{array}{rl}& \\ y& ={\displaystyle \frac{U}{V}\to y{}^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\\ & \end{array}\\ \begin{array}{rl}f{}^{\prime }(x)& =-{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{3}{2}}}}\\ & =-{\displaystyle \frac{1}{2x^{{}^{\frac{3}{2}}}}}\\ & =-{\displaystyle \frac{1}{2x^1.x^{{}^{\frac{1}{2}}}}}\\ & =-{\displaystyle \frac{1}{2x\sqrt{x}}}\end{array}& \begin{array}{rl}f{}^{\prime }(x)& ={\displaystyle \frac{0.\sqrt{x}-1.\frac{1}{2}{x}^{{}^{\frac{1}{2}-1}}}{{\left(\sqrt{x}\right)}^2}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}}{x}}\\ & =-{\displaystyle \frac{1}{2x{x}^{{}^{\frac{1}{2}}}}}\\ & =-{\displaystyle \frac{1}{2x\sqrt{x}}}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Turunan pertama dari}y={\displaystyle \sqrt[n]{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{n}{x}^{{}^{\frac{1}{n}}}}& & \text{d}.(n-1)\sqrt[n-1]{x}\\ \text{b}.{\displaystyle \frac{1}{n}{x}^{{}^{\frac{1-n}{n}}}}& \text{c}.{\displaystyle \frac{1}{n-1}{x}^{{}^{n-1}}}& \text{e}.\sqrt[n-1]{x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}y& =\sqrt[n]{x}={\displaystyle x^{{}^{\frac{1}{n}}},\text{maka}}\\ y{}^{\prime }& ={\displaystyle \frac{1}{n}{x}^{{}^{\frac{1}{n}-1}}={\displaystyle \frac{1}{n}{x}^{{}^{\frac{1-n}{n}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Turunan ke}-n\text{dari}y={\displaystyle \frac{1}{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle n!.x^{-(n+1)}}& \\ \text{b}.{\displaystyle (n+1)!.x^{-(n+1)}}& \\ \text{c}.{\displaystyle (-1{)}^{n}n!.x^{-(n+1)}}\\ \text{d}.(-1{)}^{n+1}n!.x^{-(n+1)}\\ \text{e}.(-1{)}^{n+1}(n+1)!.x^{-(n+1)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ccl|}\hline \text{Fungsi}& y={\displaystyle \frac{1}{x}}& y=x^{-1}\\ y^{\prime }={\displaystyle \frac{dy}{dx}}& -x^{-2}& (-1{)}^1.1!.x^{-(1+1)}\\ y^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}}& 2x^{-3}& (-1{)}^2.2!.x^{-(2+1)}\\ y^{‴}={\displaystyle \frac{d^{3}y}{d{x}^3}}& -6x^{-4}& (-1{)}^3.3!.x^{-(3+1)}\\ y^{IV}={\displaystyle \frac{d^{4}y}{d{x}^4}}& 24x^{-5}& (-1{)}^4.4!.x^{-(4+1)}\\ y^V={\displaystyle \frac{d^{5}y}{d{x}^5}}& \cdots & \cdots \\ ⋮& ⋮& ⋮\\ y^n={\displaystyle \frac{d^{n}y}{d{x}^n}}& \cdots & (-1{)}^n.n!.x^{-(n+1)}\\ \hline\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 5.& \text{Diketahui jumlahdua bilangan adalah 30}\\ & \text{Jika perkalian salah satu bilangan dengan}\\ & \text{kuadrat bilangan lainnya mencapai nilai}\\ & \text{maksimum, tentukanlah}:\\ & \text{a}.\text{kedua bilangan tersebut}\\ & \text{b}.\text{nilai maksimumnya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan salah satu bilangan adalah}x,\\ & \text{maka bilangan yang lainnya adalah}:30-x\\ & \text{Dan misalkan pula perkalian ini dirumuskan}\\ & \text{dengan}p(x),\text{maka}\\ & \begin{array}{rl}\text{a}.p(x)& =(30-x)x^2=30x^2-x^3\\ \text{syara}& \text{t ekstrim maksimum}{p}^{\prime }(x)=0\\ 60x-& 3x^2=0\iff 3x(20-x)=0\\ x_1=& 0\text{atau}{x}_2=20\\ \text{Jadi},& \text{nilai agar maksimum}x=20\text{dan}\\ \text{nilai}& x\text{yang lain adalah}:30-20=10\\ \text{b}.\text{Dan}& \text{nilai maksimumnya adalah}:\\ p(20)& =(30-20){.20}^2=10.400=4000(\mathbf{\text{mak}})\end{array}\end{array}$ .

$\begin{array}{l}\\ 6.& \text{Sebuah peluru ditembakkan ke atas. Dan}\\ & \text{tinggi yang dapat dicapai peluru adalah}\\ & h\text{meter dalam waktu}t\text{detik yang dapat}\\ & \text{rumuskan dengan}h(t)=120t-5t^2\\ & \text{Tentukanlah}\\ & \text{a}.t\text{agar}h\text{maksimum}\\ & \text{b}.\text{tinggi}h\text{maksimum}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.\text{syara}& \text{at ekstrim maksimum}{h}^{\prime }(x)=0\\ 120-& 10t=0\iff t=12\\ \text{Jadi},& \text{tinggi maksimum dicapai saat}t=12\\ \text{b}.\text{Dan}& \text{tinggi maksimumnya adalah}:\\ h(12)& =120(12)-5(12{)}^2=720\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& (\mathbf{\text{OSK 2018}})\\ & \text{Diketahui bilangan real}x\text{dan}y\\ & \text{yang memenuhi}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Nilai minimum}{\displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\text{adalah}....}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misal}t={\displaystyle \frac{x}{y}}\\ & \text{Misalkan juga}f(t)={\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}\\ & \text{maka}f(t)={\displaystyle \frac{2t^2-3t+4}{-2t^2+5t-2}}\\ & \bullet \text{Agar minimum, maka}{f}^{\prime }(t)=0\\ & \text{Sehingga}\\ & f^{\prime }(t)=4t^2+8t-14=0\\ & \iff t_{1,2}=-1\pm {\displaystyle \frac{3}{2}\sqrt{2}}\\ & \text{Pilih yang positif, yaitu}t=-1+{\displaystyle \frac{3}{2}\sqrt{2}}\\ & \bullet \text{Dengan proses substistusi harga}t\\ & \text{di atas, maka akan didapatkan }\\ & \text{nilai}f(t)=1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Menurut bentuk}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{jelas bahwa baik}2y-x\text{dan}2x-y\\ & \text{keduanya}\mathbf{\text{positif}}\\ & \text{Lihat tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Bentuk}& \text{Pengecekan 1}& \text{Pengecekan 2}\\ \begin{array}{rl}& {\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Jelas bahwa}\\ & x,y\ne 0\\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat}(\times y)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(y)<\frac{x}{y}(y)<2(y)}\\ & \iff {\displaystyle \frac{1}{2}y<x<2y}\\ & \text{Jelas bahwa}\\ & 2y-x>0\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat dibali posisinya}\\ & {\displaystyle \frac{1}{2}<\frac{y}{x}<2}\\ & \text{Saat}(\times x)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(x)<\frac{y}{x}(x)<2(x)}\\ & \iff {\displaystyle \frac{1}{2}x<y<2x}\\ & \text{Jelas bahwa}\\ & 2x-y>0\end{array}\\ \hline\end{array}\\ & \text{Saat masing-masing}\\ & \bullet {\displaystyle \frac{x}{2y-x}={\displaystyle \frac{1}{3}+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\text{dan}}}\\ & \bullet {\displaystyle \frac{2y}{2x-y}={\displaystyle \frac{2}{3}+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}}\\ & \text{Dengan ketaksamaan AM-GM diperoleh}\\ & \begin{array}{rl}{\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)\\ & \ge 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{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Jika}x,y,z\text{adalah sisi sebuah segitiga}\\ & \text{Tunjukkan bahwa nilai minimum}\\ & {\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\text{adalah 3}}\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Misalkan}{\displaystyle \frac{y}{x}=p,\frac{z}{y}=q,\text{dan}{\displaystyle \frac{z}{x}=pq,}}\\ & \text{maka soal dapat kita ubah menjadi}\\ & {\displaystyle \frac{1}{p}+\frac{1}{q}+pq}\\ & \text{Kita tahu bahwa}{(\sqrt{{\displaystyle \frac{1}{p}}}-\sqrt{{\displaystyle \frac{1}{q}}})}^2\ge 0\\ & \iff {\displaystyle \frac{1}{p}+\frac{1}{q}\ge 2\sqrt{{\displaystyle \frac{1}{pq}}}}\\ & \text{Kita sesuaikan soal, yaitu}\\ & {\displaystyle \frac{1}{p}+\frac{1}{q}+pq\ge pq+2\sqrt{{\displaystyle \frac{1}{pq}}}}\\ & \text{Misalkan}\sqrt{{\displaystyle \frac{1}{pq}}}=a,{\displaystyle \frac{1}{pq}=a^2,pq={\displaystyle \frac{1}{a^2}}}\\ & \text{dengan}a\ge 0,\text{maka}\\ & {\displaystyle \frac{1}{p}+\frac{1}{q}+pq={\displaystyle \frac{1}{a^2}+2a=f(a)}}\\ & \text{Syarat ekstrim minimum}:f^{\prime }(a)=0\\ & -2a^{-3}+2=0\iff 2=2a^{-3}\iff 1=a^{-3}\\ & \iff 1={\displaystyle \frac{1}{a^3}\iff a^3=1\iff a=1}\\ & \text{Sehingga nilai minimumnya saat}a=1\\ & \text{dengan nilai}f(1)={\displaystyle \frac{1}{1^2}+2.1=3}\\ & \text{Jadi, nilai nilai minimum}{\displaystyle \frac{1}{p}+\frac{1}{q}+pq=3}\\ & \text{atau}{\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\text{adalah 3 atau juga}}\\ & \text{dapat dituliskan dengan}{\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\ge 3}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\\ & {\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\ge 3\sqrt[3]{{\displaystyle \frac{x}{y}.\frac{y}{x}.\frac{x}{z}}}=3.\sqrt[3]{1}=3}\end{array}$

DAFTAR PUSTAKA

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

SUMBER WEBSITE

1. 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

$\begin{array}{|cll|}\hline \text{No}& \text{Turunan Pertama}& \text{Turunan Pertama}\\ 1.& \text{Gradien garis singgung}& \begin{array}{rl}m& =f{}^{\prime }(x)\\ & =\underset{h\to 0}{\text{Lim}}={\displaystyle \frac{f(x+h)-f(x)}{h}}\end{array}\\ 2.& \text{Fungsi naik dan turun}& y=f(x)\{\begin{array}{ll}{f}^{\prime }(x)>0,& \\ \text{ fungsi naik }& \\ \\ f^{\prime }(x)<0,& \\ \text{ fungsi turun }& \end{array}\\ 3.& \text{Jarak, kec, percepatan}& y=s(t)\{\begin{array}{ll}s(t)& \text{ jarak}\\ s{}^{\prime }(t)& \text{ kecepatan }\\ s{}^{″}(t)& \text{ percepatan}\end{array}\\ 4.\text{a}& \text{Stasioner}& \begin{array}{rl}\text{Maksimum}:& \\ \to f& {}^{″}(k)<0\\ \text{titik mak}& (k,f(k))\end{array}\\ 4.\text{b}& \text{Stasioner}& \begin{array}{rl}\text{Minimum}:& \\ \to f& {}^{″}(k)>0\\ \text{titik min}& (k,f(k))\end{array}\\ 4.\text{c}& \begin{array}{rl}& \text{Syarat stasioner}\\ & f^{\prime }(x)=0,\to x\\ & \text{dengan}x=k\end{array}& \begin{array}{rl}\text{Belok}:& \\ \to f& {}^{″}(k)=0\\ \text{titik belok}& (k,f(k))\end{array}\\ 5.& \begin{array}{rl}& \text{Limit fungsi}\\ & \text{bentuk tak tentu}\end{array}& \begin{array}{rl}& \text{Aturan L'Hopital}\\ & \\ & \underset{x\to h}{\text{Lim}}{\displaystyle \frac{f(x)}{g(x)}=\underset{x\to h}{\text{Lim}}{\displaystyle \frac{f{}^{\prime }(x)}{g{}^{\prime }(x)}}}\\ & \\ & \text{untuk hasil limit}\\ & \text{bentuk}\frac{0}{0}\text{atau}\frac{\mathrm{\infty }}{\mathrm{\infty }}\end{array}\\ \text{No}& \text{Turunan Kedua}& \text{Turunan Kedua}\\ 6.& f^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}}& \begin{array}{rl}\bullet & \text{Belok}\\ \bullet & \text{Percepatan}\\ \bullet & \text{Maksimum}\\ \bullet & \text{Minimum}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

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

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

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai stasioner fungsi}\\ & f(x)=x^3+3x^2-9x+5\text{dan }\\ & \text{tentukan pula jenisnya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahu fungsi}\\ & f(x)=x^3+3x^2-9x+5\\ & \text{Syarat stasioner}{f}^{\prime }(x)=0,\text{maka}\\ & f{}^{\prime }(x)=3x^2+6x-9=0\\ & 0=3(x+3)(x-1)\\ & x=-3\text{atau}x=1\\ & \text{Dan untuk nilai dan titik stasionernya}:\\ & f(-3)=(-3{)}^3+3(-3{)}^2-9(-3)+5=32\\ & \to (-3,32)\text{adalah titik balik maksimum}\\ \\ & f(1)=(1{)}^3+3(1{)}^2-9(1)+5=0\\ & \to (1,0)\text{adalah titik balik minimum}\\ \\ & \text{Dan berikut ilustrasi gambarnya}\\ & \text{untuk soal no.2 dan 3}\end{array}\end{array}$.

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

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

 Aturan Rantai

$\begin{array}{rl}& \\ \text{Untuk}:& \\ & \{\begin{array}{ll}a& \in \mathbb{R}\\ n& \in \mathbb{Q}\\ c& \text{konstanta}\\ U& =g(x)\\ V& =h(x)\end{array}\\ & \end{array}$.

$\begin{array}{|l|}\hline \mathbf{\text{Sifat-Sifat}}\\ \begin{array}{rl}y& =c\to y^{\prime }=0\\ y& =c.U\to y^{\prime }=c.U^{\prime }\\ y& =U\pm V\to y^{\prime }=U^{\prime }\pm V^{\prime }\\ y& =U.V\to y^{\prime }=U^{\prime }.V+U.V^{\prime }\\ y& ={\displaystyle \frac{U}{V}\to y^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\end{array}\\ \mathbf{\text{Fungsi Aljabar}}\\ \begin{array}{rl}y& =a.x^n\to y{}^{\prime }=n.a.x^{(n-1)}\\ y& =a.U^n\to y{}^{\prime }=n.a.U^{(n-1)}.U^{\prime }\end{array}\\ \mathbf{\text{Fungsi Trigonometri}}\\ \begin{array}{rlrl}y& =a\sin U\to & y^{\prime }& =(a\cos U).U^{\prime }\\ y& =a\cos U\to & y^{\prime }& =(-a\sin U).U^{\prime }\\ y& =a\tan U\to & y^{\prime }& =(a{\sec }^{2}U).U^{\prime }\end{array}\\ \mathbf{\text{Aturan rantai }}\\ \begin{array}{rl}& \text{pada turunan untuk}y=f(u),\text{jika}\\ & \text{untuk}u\text{merupakan fungsi}x,\text{maka}:\\ & y{}^{\prime }=f{}^{\prime }(x).u^{\prime }\text{atau}{\displaystyle \frac{dy}{dx}={\displaystyle \frac{dy}{du}.\frac{du}{dx}}}\\ & \end{array}\\ \hline\end{array}$.

Turunan Kedua Fungsi Aljabar

$\begin{array}{rl}& \\ \text{Suatu fungsi}& \\ & f:x\to y\\ & \end{array}$.

$\begin{array}{|cc|}\hline \text{Notasi}& \text{Proses}\\ \begin{array}{rl}& \\ & \{\begin{array}{ll}f{}^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ \\ f{}^{″}(x)& ={\displaystyle \frac{d^{2}y}{d{x}^2}}\end{array}\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \\ y& =a{x}^n\\ y^{\prime }& =n.a.x^{n-1}\\ y^{″}& =(n-1).n.a.x^{n-2}\\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai dari}{\displaystyle \frac{dy}{dx}}\\ & \text{a}.y=\sqrt{x}\\ & \text{b}.y=\sqrt{3+\sqrt{x}}\\ & \text{c}.y=\sqrt{3+\sqrt{3+\sqrt{x}}}\\ & \text{d}.y=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{x}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\text{a}.y& =\sqrt{x}\\ & =x^{{}^{\frac{1}{2}}}\\ y^{\prime }& ={\displaystyle \frac{1}{2}.x^{{}^{\frac{1}{2}-1}}}\\ & ={\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{2x^{{}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{1}{2\sqrt{x}}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{b}.y& =\sqrt{3+\sqrt{x}}\\ & ={(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}}}\\ y^{\prime }& ={\displaystyle \frac{1}{2}{(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}-1}}.{\displaystyle \frac{1}{2}{x}^{{}^{\frac{1}{2}-1}}}}\\ & ={\displaystyle \frac{1}{4}{(3+x^{{}^{\frac{1}{2}}})}^{{}^{-\frac{1}{2}}}.x^{{}^{-\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{4{(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}}}.x^{{}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{1}{4\sqrt{3+\sqrt{x}}.\sqrt{x}}}\\ & \end{array}\\ \hline\end{array}\end{array}$.

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

$\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ 2.\text{a}& \text{Alternatif 1}\\ & f(x)=2\sin x\cos x\\ & \begin{array}{rl}f(x)& =\sin 2x\\ f{}^{\prime }(x)& =\cos 2x.(2)\\ & =2\cos 2x\end{array}\\ \\ & \text{Alternatif 2}\\ & f(x)=y=\sin 2x\\ & \{\begin{array}{ll}y=\sin u& \to {\displaystyle \frac{dy}{du}=\cos u}\\ \\ u=2x& \to {\displaystyle \frac{du}{dx}=2}\end{array}\\ & \text{sehingga}\\ & \begin{array}{rl}y& =\sin 2x\\ {\displaystyle \frac{dy}{dx}}& ={\displaystyle \frac{dy}{du}.\frac{du}{dx}}\\ & =\cos u.2\\ & =2\cos u\\ & =2\cos 2x\end{array}\end{array}\end{array}$. 

$\begin{array}{l}\\ 2.\text{b}& f(x)=x^2\sin x\\ & \begin{array}{rl}f{}^{\prime }(x)& =2x^{2-1}.(\sin x)+x^2.(\cos x)\\ & =2x\sin x+x^2\cos x\end{array}\\ \text{c}& f(x)=3{\cos }^{2}x\\ & f^{\prime }(x)=2.3.{\cos }^{2-1}x.(-\sin x)\\ & =-6\sin x\cos x\\ & =-3\sin 2x\\ \text{d}& f(x)=3{\sin }^{2}x-x^3\\ & f^{\prime }(x)=2.3.{\sin }^{2-1}x.(\cos x)-3x^{3-1}\\ & =6\sin x\cos x-3x^2\\ & =3\sin 2x-3x^2\\ \text{e}& f(x)=5{\sin }^{4}x\\ & f^{\prime }(x)=4.5.{\sin }^{4-1}x.(\cos x)\\ & =20{\sin }^{3}x\cos x\\ & \text{atau boleh juga}\\ & =20{\sin }^{2}x\sin x\cos x\\ & =20{\sin }^{2}x\sin 2x=20\sin 2x{\sin }^{2}x\end{array}$ .

$\begin{array}{l}\\ 2.\text{m}& y=f(x)={\sin }^2(\pi -3x)\\ & \begin{array}{rl}f{}^{\prime }(x)& =2{\sin }^{2-1}(\pi -3x).\cos (\pi -3x).(-3)\\ & =-6\sin (\pi -3x)\cos (\pi -3x)\\ & =-3.2\sin (\pi -3x)\cos (\pi -3x)\\ & =-3\sin 2(\pi -3x)\\ & =-3\sin (2\pi -6x)\end{array}\\ \\ & \text{atau boleh juga}\\ & \begin{array}{rl}y=& f(x)={\sin }^2(\pi -3x)\\ & \{\begin{array}{ll}y=u^2& \to {\displaystyle \frac{dy}{du}=2u.}\\ \\ u=\sin w& \to {\displaystyle \frac{du}{dw}=\cos w}\\ \\ w=\pi -3x& \to {\displaystyle \frac{dw}{dx}=-3}\end{array}\\ {\displaystyle \frac{dy}{dx}}& ={\displaystyle \frac{dy}{du}.\frac{du}{dw}.\frac{dw}{dx}}\\ & =\left(2u\right).(\cos w).(-3)\\ & =-3.(2\sin w).(\cos w)\\ & =-3\sin 2w\\ & =-3\sin (2\pi -6x)\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai dari}{\displaystyle \frac{d^{2}y}{d{x}^2}}\\ & \text{a}.y=\sqrt{x}\\ & \text{b}.y=\sqrt{3+\sqrt{x}}\\ & \text{c}.y=15x^3-4x\\ & \text{d}.y={\displaystyle \frac{x^4}{3}+\frac{x^2}{2}+x+\sqrt{x}+1+{\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^2}+\frac{3}{x^4}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Lihat pembahasan soal no.1, yaitu }\\ & y=\sqrt{x}\\ & y^{\prime }={\displaystyle \frac{1}{2\sqrt{x}}}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}3.\text{a}.y& =\sqrt{x}\\ y^{\prime }& ={\displaystyle \frac{dy}{dx}={\displaystyle \frac{1}{2\sqrt{x}}={\displaystyle \frac{1}{2}{x}^{-\frac{1}{2}}}}}\\ y^{″}& ={\displaystyle \frac{d^{2}y}{d{x}^2}=(-{\displaystyle \frac{1}{2}}){\displaystyle \frac{1}{2}.x^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{4x^{\frac{1}{2}+1}}=-{\displaystyle \frac{1}{4x\sqrt{x}}}}\end{array}\end{array}$.

$.\begin{array}{|l|}\hline \begin{array}{rl}& 3.\text{d}\\ & \begin{array}{rl}y& ={\displaystyle \frac{x^4}{3}+\frac{x^2}{2}+x+\sqrt{x}+1+{\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^2}+\frac{3}{x^4}}}\\ & \parallel \\ & \mathbf{\text{Turunan pertama}}\\ & \Downarrow \\ {\displaystyle \frac{dy}{dx}=y{}^{\prime }}& ={\displaystyle \frac{4x^3}{3}+x+1+{\displaystyle \frac{1}{2\sqrt{x}}+0-\frac{1}{2x\sqrt{x}}-\frac{1}{x^2}-\frac{2}{x^3}-\frac{12}{x^5}}}\\ & ={\displaystyle \frac{4x^3}{3}+x+1+{\displaystyle \frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}-\frac{1}{x^2}-\frac{2}{x^3}-\frac{12}{x^5}}}\\ & \parallel \\ & \mathbf{\text{Turunan kedua}}\\ & \Downarrow \\ {\displaystyle \frac{d^{2}y}{d{x}^2}=y{}^{″}}& =4x^2+1+0-{\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}+\frac{2}{x^3}+\frac{6}{x^4}+\frac{60}{x^6}}\\ & =4x^2+1-{\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}+\frac{2}{x^3}+\frac{6}{x^4}+\frac{60}{x^6}}\end{array}\\ & \end{array}\\ \hline\end{array}$.

$\text{LATIHAN SOAL}$.

Silahkan kerjakan soal yang belum diselesaikan atau dijawab

DAFTAR PUSTAKA

1. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU
2. Kartrini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: ESIS
4. 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

$\begin{array}{rl}& \\ \text{Untuk}:& \\ & \{\begin{array}{ll}a& \in \mathbb{R}\\ n& \in \mathbb{Q}\\ c& \text{konstanta}\\ U& =g(x)\\ V& =h(x)\end{array}\\ & \end{array}$.

$\begin{array}{|l|}\hline \mathbf{\text{Sifat-Sifat}}\\ \begin{array}{rl}y& =c\to y^{\prime }=0\\ y& =c.U\to y^{\prime }=c.U^{\prime }\\ y& =U\pm V\to y^{\prime }=U^{\prime }\pm V^{\prime }\\ y& =U.V\to y^{\prime }=U^{\prime }.V+U.V^{\prime }\\ y& ={\displaystyle \frac{U}{V}\to y^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\end{array}\\ \mathbf{\text{Fungsi Aljabar}}\\ \begin{array}{rl}y& =a.x^n\to y{}^{\prime }=n.a.x^{(n-1)}\\ y& =a.U^n\to y{}^{\prime }=n.a.U^{(n-1)}.U^{\prime }\end{array}\\ \mathbf{\text{Fungsi Trigonometri}}\\ \begin{array}{rlrl}y& =a\sin U\to & y^{\prime }& =(a\cos U).U^{\prime }\\ y& =a\cos U\to & y^{\prime }& =(-a\sin U).U^{\prime }\\ y& =a\tan U\to & y^{\prime }& =(a{\sec }^{2}U).U^{\prime }\end{array}\\ \mathbf{\text{Aturan rantai }}\\ \begin{array}{rl}& \text{pada turunan untuk}y=f(u),\text{jika}\\ & \text{untuk}u\text{merupakan fungsi}x,\text{maka}:\\ & y{}^{\prime }=f{}^{\prime }(x).u^{\prime }\text{atau}{\displaystyle \frac{dy}{dx}={\displaystyle \frac{dy}{du}.\frac{du}{dx}}}\\ & \end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

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

$\begin{array}{rl}& \mathbf{\text{Bukti}}\\ & \begin{array}{l}\\ 1.\text{a}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a(x+h{)}^n-a{x}^n}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a(x^n+(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}.h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^3+\cdots +(\genfrac{}{}{0ex}{}{n}{n-1})x.h^{n-1}+h^n)-a{x}^n}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{a((\genfrac{}{}{0ex}{}{n}{1})x^{n-1}.h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^3+\cdots +(\genfrac{}{}{0ex}{}{n}{n-1})x.h^{n-1}+h^n)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{ah((\genfrac{}{}{0ex}{}{n}{1})x^{n-1}+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^2+\cdots +(\genfrac{}{}{0ex}{}{n}{n-1})x.h^{n-2}+h^{n-1})}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle a(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}+a(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}.h+a(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}.h^2+\cdots +a{h}^{n-1}}\\ & =a(\genfrac{}{}{0ex}{}{n}{1})x^{n-1}+0+0+...+0\\ & =a.{\displaystyle \frac{n!}{(n-1)!.1!}{x}^{n-1}}\\ & =a.{\displaystyle \frac{n.(n-1)!}{(n-1)!}{x}^{n-1}}\\ & =a.n.x^{n-1}◼\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{b}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(u(x+h)+v(x+h))-(u(x)+v(x))}{h}}\\ & =\underset{h\to 0}{\text{Lim}}\left({\displaystyle \frac{u(x+h)-u(x)}{h}+\frac{v(x+h)-v(x)}{h}}\right)\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{u(x+h)-u(x)}{h}+\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{v(x+h)-v(x)}{h}}}\\ & =u^{\prime }(x)+v^{\prime }(x)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{c}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(u(x+h)\times v(x+h))-(u(x)\times v(x))}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{u(x+h)\times v(x+h)-u(x+h)\times v(x)+u(x+h)\times v(x)-u(x)\times v(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}(u(x+h)\times {\displaystyle \frac{v(x+h)-v(x)}{h}+v(x)\times {\displaystyle \frac{u(x+h)-u(x)}{h}}})\\ & =\underset{h\to 0}{\text{Lim}}u(x+h)\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{v(x+h)-v(x)}{h}+\underset{h\to 0}{\text{Lim}}v(x)\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{u(x+h)-u(x)}{h}}}\\ & =u(x)\times v^{\prime }(x)+v(x)\times u^{\prime }(x)\\ & =u^{\prime }(x)\times v(x)+u(x)\times v^{\prime }(x)◼\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{d}& \text{Misalkan}p(x)={\displaystyle \frac{u(x)}{v(x)}}\\ & \text{Sebelumnya telah diketahui dari no. 1. c}\\ & u(x)=p(x)\times v(x)\\ & \begin{array}{rl}u{}^{\prime }(x)& =p^{\prime }(x)\times v(x)+p(x)\times v^{\prime }(x)\\ \text{Sekar}& \text{ang kita substitusikan pemisalan}\\ \text{di at}& \text{as, yaitu}:\end{array}\\ & \begin{array}{rl}{p}^{\prime }(x)& \times v(x)=u^{\prime }(x)-p(x)\times v^{\prime }(x)\\ & =u^{\prime }(x)-{\displaystyle \frac{u(x)}{v(x)}\times v^{\prime }(x)}\\ & ={\displaystyle \frac{u^{\prime }(x)\times v(x)-u(x)\times v^{\prime }(x)}{v(x)}}\\ p^{\prime }(x)& ={\displaystyle \frac{u^{\prime }(x)\times v(x)-u(x)\times v^{\prime }(x)}{v^2(x)}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{e}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2x+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2x+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2x+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2x)}\\ & =\cos x◼\end{array}\end{array}$ .

$\begin{array}{l}\\ 1.\text{f}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos (x+h)-\cos x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2\sin {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h).\frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h)\times \frac{1}{2}}}\\ & =-2\sin {\displaystyle \frac{1}{2}(2x+0)\times \frac{1}{2}}\\ & =-\sin {\displaystyle \frac{1}{2}(2x)}\\ & =-\sin x◼\end{array}\end{array}$.

$\begin{array}{l}\\ 1.\text{g}& f{}^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \begin{array}{rl}f{}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\tan x+\tan h}{1-\tan x.\tan h}-\tan x}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\tan x+\tan h-\tan x+{\tan }^{2}x.\tan h}{1-\tan x.\tan h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan h(1+{\tan }^{2}x)}{h(1-\tan x.\tan h)}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan h}{h}\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{1+{\tan }^{2}x}{1-\tan x.\tan h}}}\\ & =1\times {\displaystyle \frac{1+{\tan }^{2}x}{1-0}}\\ & =1+{\tan }^{2}x\\ & ={\sec }^{2}x◼\end{array}\end{array}$.

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

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

$\begin{array}{|cc|}\hline \text{g}& \text{h}\\ \begin{array}{rl}y& =(x-3{)}^2\\ y{}^{\prime }& =2.(x-3{)}^{2-1}.1\\ & =2(x-3)\\ & \end{array}& \begin{array}{rl}y& ={(x^3-2)}^2\\ y{}^{\prime }& =2.{(x^3-2)}^{2-1}.3x^2\\ & =6(x^3-2)x^2\\ & =6x^5-12x^2\end{array}\\ \text{i}& \text{j}\\ \begin{array}{rl}y& =(x+1)(x-2)\\ y{}^{\prime }& =1.(x+2)+(x+1).1\\ & =2x+3\end{array}& \begin{array}{rl}y& =(3-x)(5-x)\\ y{}^{\prime }& =-1.(5-x)+(3-x).-1\\ & =2x-8\end{array}\\ \hline\end{array}$ .

$\begin{array}{|cccc|}\hline \text{k}& \text{l}& {\text{m}}_1& {\text{m}}_2\\ \begin{array}{rl}y& =(2x+3{)}^2\\ y{}^{\prime }& =2.(2x+3{)}^{2-1}.2\\ & =4(2x+3{)}^1\\ & =8x+12\\ & \\ & \end{array}& \begin{array}{rl}y& =(x-2{)}^3\\ y{}^{\prime }& =3(x-2{)}^{3-1}.1\\ & =3(x-2{)}^2\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& =(4x+1)(4x-1)\\ \text{c}& \text{ara 1}\\ y{}^{\prime }& =4(4x-1)+(4x+1).4\\ & =16x-4+16x+4\\ & =32x\\ & \end{array}& \begin{array}{rl}y& =(4x+1)(4x-1)\\ \text{c}& \text{ara 2}\\ y& =16x^2-1\\ y{}^{\prime }& =2.16x^{2-1}-0\\ & =32x^1\\ & =32x\end{array}\\ \text{n}& \text{o}& \text{p}& \text{q}\\ \begin{array}{rl}y& =(x-1)(x+1)(x+2)\\ & =(x^2-1)(x+2)\\ & =x^3+2x^2-x-2\\ y{}^{\prime }& =3x^{3-1}+2.2x^{2-1}-x^{1-1}-0\\ & =3x^2+4x-1\\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{1}{2}{x}^{-4}}\\ y{}^{\prime }& =-4.{\displaystyle \frac{1}{2}{x}^{-4-1}}\\ & =-2x^{-5}\\ & =-{\displaystyle \frac{2}{x^5}}\\ & \end{array}& \begin{array}{rl}y& =x^{-5}\\ y^{\prime }& =-5.x^{-5-1}\\ & =-5x^{-6}\\ & =-{\displaystyle \frac{5}{x^6}}\\ & \\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{2}{x^3}}\\ & =2x^{-3}\\ y{}^{\prime }& =-3.2x^{-3-1}\\ & =-6x^{-4}\\ & =-{\displaystyle \frac{6}{x^4}}\end{array}\\ \hline\end{array}$.

$\begin{array}{|cccc|}\hline \text{r}& \text{s}& \text{t}& \text{u}\\ \begin{array}{rl}y& ={\displaystyle \frac{1}{2\sqrt{x}}}\\ & ={\displaystyle \frac{1}{2x^{\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\\ y{}^{\prime }& =-{\displaystyle \frac{1}{2}.\frac{1}{2}{x}^{{}^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{4}{x}^{{}^{-\frac{3}{2}}}}\\ & =-{\displaystyle \frac{1}{4x^{{}^{\frac{3}{2}}}}}\\ & =-{\displaystyle \frac{1}{4}\sqrt{x^3}}\end{array}& \begin{array}{rl}y& ={\displaystyle \frac{1}{3x^5}}\\ & ={\displaystyle \frac{1}{3}{x}^{-5}}\\ y{}^{\prime }& =-5.{\displaystyle \frac{1}{3}{x}^{-5-1}}\\ & =-{\displaystyle \frac{5}{3}{x}^{-6}}\\ & =-{\displaystyle \frac{5}{3x^6}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& =2x^4+{\displaystyle \frac{1}{2x^3}}\\ & =2x^4+{\displaystyle \frac{1}{2}{x}^{-3}}\\ y{}^{\prime }& =4.2x^{4-1}+(-3).{\displaystyle \frac{1}{2}{x}^{-3-1}}\\ & =8x^3-{\displaystyle \frac{3}{2}{x}^{-4}}\\ & =8x^3-{\displaystyle \frac{3}{2x^4}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{x}{2}+\frac{2}{x}}\\ & ={\displaystyle \frac{1}{2}x+2x^{-1}}\\ y{}^{\prime }& ={\displaystyle \frac{1}{2}{x}^{1-1}+(-1).2x^{-1-1}}\\ & ={\displaystyle \frac{1}{2}{x}^0-2x^{-2}}\\ & ={\displaystyle \frac{1}{2}-{\displaystyle \frac{2}{x^2}}}\\ & \\ & \\ & \end{array}\\ \hline\end{array}$.

$\begin{array}{|cc|}\hline \text{v}& \text{w}\\ \begin{array}{rl}y& ={(2x^3+{\displaystyle \frac{1}{x}})}^2\\ & ={(2x^3+x^{-1})}^2\\ y{}^{\prime }& =2.{(2x^3+x^{-1})}^{2-1}.(3.2x^{3-1}+(-1)x^{-1-1})\\ & =2.{(2x^3+x^{-1})}^1.(6x^2-x^{-2})\\ & =2(2x^3+{\displaystyle \frac{1}{x}})(6x^2-{\displaystyle \frac{1}{x^2}})\\ & =2(12x^5-2x+6x-{\displaystyle \frac{1}{x^3}})\\ & =24x^5+8x-{\displaystyle \frac{2}{x^3}}\\ & \end{array}& \begin{array}{rl}y& =x^2{(1+\sqrt{x})}^2\\ & =x^2{(1+x^{{}^{\frac{1}{2}}})}^2\\ y{}^{\prime }& =2x.{(1+x^{{}^{\frac{1}{2}}})}^2+x^2.2.{(1+x^{{}^{\frac{1}{2}}})}^{2-1}.(0+{\displaystyle \frac{1}{2}.x^{{}^{\frac{1}{2}-1}}})\\ & =2x{(1+\sqrt{x})}^2+2x^2.(1+\sqrt{x}).\left({\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\right)\\ & =2x{(1+\sqrt{x})}^2+2x^2.\left({\displaystyle \frac{1}{2x^{{}^{\frac{1}{2}}}}}\right).(1+\sqrt{x})\\ & =2x{(1+\sqrt{x})}^2+x^{{}^{2-\frac{1}{2}}}.(1+\sqrt{x})\\ & =2x{(1+\sqrt{x})}^2+x^{{}^{\frac{3}{2}}}(1+\sqrt{x})\\ & =2x{(1+\sqrt{x})}^2+(x\sqrt{x}+x^2)\end{array}\\ \hline\end{array}$.

$\begin{array}{|cc|}\hline {\text{x}}_1& {\text{x}}_2\\ \begin{array}{rl}y& ={\displaystyle \frac{U}{V}}\\ y{}^{\prime }& ={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}\end{array}& \begin{array}{rl}y& =UV\\ y{}^{\prime }& =U^{\prime }.V+U.V^{\prime }\end{array}\\ \begin{array}{rlrl}& & & \\ U& =2+4x& \to U^{\prime }& =4\\ V& =x^2& \to V^{\prime }& =2x\\ & & & \end{array}& \begin{array}{rlrl}& & & \\ U& =2+4x& \to U^{\prime }& =4\\ V& =x^2& \to V^{\prime }& =2x\\ & & & \end{array}\\ \begin{array}{rl}y& ={\displaystyle \frac{2+4x}{x^2}}\\ y{}^{\prime }& ={\displaystyle \frac{(4)(x^2)-(2+4x).(2x)}{{\left(x^2\right)}^2}}\\ & ={\displaystyle \frac{4x^2-4x-8x^2}{x^4}}\\ & ={\displaystyle \frac{-4x^2-4x}{x^4}}\\ & ={\displaystyle \frac{-4x-4}{x^3}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}y& ={\displaystyle \frac{2+4x}{x^2}}\\ & =(2+4x).x^{-2}\\ y{}^{\prime }& =(4).x^{-2}+(2+4x).-2x^{-2-1}\\ & =4x^{-2}-(4+8x).x^{-3}\\ & =4x^{-2}-4x^{-3}-8x^{-2}\\ & =-4x^{-3}-4x^{-2}\\ & ={\displaystyle \frac{-4}{x^3}+{\displaystyle \frac{-4}{x^2}}}\\ & ={\displaystyle \frac{-4-4x}{x^3}}\\ & ={\displaystyle \frac{-4x-4}{x^3}}\end{array}\\ \hline\end{array}$.

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  $\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$  ada, maka bentuk  $\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$  disebut sebagai turunan dari fungsi  $f(x)$  saat  $x=k$.

A. 3 Notasi

• Notasi turunan fungsi dilambangkan dengan  $f^{\prime }(k)$  dengan  $f^{\prime }(k)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(k+h)-f(k)}{h}}$.
• Lambang   $f^{\prime }(k)$  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  $f^{\prime }(k)$.
• Misalkan fungsi  $f(x)$  mempunyai turunan  $f^{\prime }(x)$. Jika  $f^{\prime }(k)$  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

$y{}^{\prime }={\displaystyle \frac{dy}{dx}={\displaystyle \frac{d(f(x))}{dx}=f^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}}}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}f(x)=2x,\text{hitunglah laju }\\ & \text{perubahan fungsi}f\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}f(x)=2x\\ & \begin{array}{|c|}\hline \text{Cara Pertama}\\ \begin{array}{rl}f(x)& =2x\\ f(2)& =2.2=4\\ f^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{f(x)-f(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(2x)-(4)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{2x-4}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 2}\\ & =2\end{array}\\ \text{Cara Kedua}\\ \begin{array}{rl}f(2)& =4\\ f(2+& h)=2(2+h)=4+2h\\ f(2+& h)-f(2)=2h\\ f^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\\ & =2\end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}f\text{di}\\ & x=2\text{adalah}2\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}f(x)=3x-5,\text{hitunglah laju }\\ & \text{perubahan fungsi}f\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}f(x)=3x-5\\ & \begin{array}{|c|}\hline \text{Cara Pertama}\\ \begin{array}{rl}f(x)& =3x-5\\ f(2)& =3.2-5=1\\ f^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{f(x)-f(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(3x-5)-(1)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{3x-6}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 3}\\ & =3\end{array}\\ \text{Cara Kedua}\\ \begin{array}{rl}f(2)& =1\\ f(2+& h)=3(2+h)-5=3h+1\\ f(2+& h)-f(2)=3h\\ f^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(2+h)-f(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{3h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}3\\ & =3\end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}f\text{di}\\ & x=2\text{adalah}3\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui}f(x)=2022x^2,\text{tentukanlah}{f}^{\prime }(x)\text{dan}{f}^{\prime }(1)\\ & \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline f^{\prime }(x)& f^{\prime }(1)\\ \begin{array}{rl}f(x)& =2022x^2\\ f(x+h)& =2022(x+h{)}^2\\ & =2022(x^2+2xh+h^2)\\ & =2022x^2+4044xh+2022h^2\\ f^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(2022x^2+4044xh+2022h^2)-\left(2022x^2\right)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4044xh+2022h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 4044x+2022h}\\ & =4044x\end{array}& \begin{array}{rl}{f}^{\prime }(x)& =4044x\\ \text{maka},& \\ f^{\prime }(1)& =4044.1\\ & =4044\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa fungsi}f(x)={\displaystyle \frac{1}{x^2}\text{dengan daerah asal}}\\ & {\text{D}}_f=\{x|x\in \mathbb{R},\text{dan}x\ne 0\}.\\ & \text{a}.\text{Tunjukkan bahwa}{f}^{\prime }(a)={\displaystyle -\frac{2}{a^3}}\\ & \text{b}.\text{jelaskanlah mengapa}{f}^{\prime }(0)\text{tidak terdefinisi}\\ & \\ & \text{Jawab}:\\ \\ & \begin{array}{|ll|}\hline \begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{(x+h{)}^2}-{\displaystyle \frac{1}{x^2}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{x^2-(x+h{)}^2}{{(x(x+h))}^2}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{x^2-(x^2+2xh+h^2)}{h{(x(x+h))}^2}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2xh-h^2}{h{(x(x+h))}^2}}\\ & =\underset{h\to 0}{\text{Lim}}-{\displaystyle \frac{2x+h}{{(x(x+h))}^2}}\\ & =-{\displaystyle \frac{2x}{x^4}}\\ & =-{\displaystyle \frac{2}{x^3}}\end{array}& \begin{array}{rl}& \text{Untuk}\text{jawaban poin a dan b }\\ & \text{adalah sebagai berikut}\\ & f^{\prime }(x)=-{\displaystyle \frac{2}{x^3}}\\ & f^{\prime }(a)=-{\displaystyle \frac{2}{a^3}}\\ & \text{maka},\\ & f^{\prime }(0)=-{\displaystyle \frac{2}{0^3}}\\ & =-{\displaystyle \frac{2}{0}}\\ & \\ & \text{karena penyebut berupa }\\ & \text{bilangan}0\\ & \text{maka}{f}^{\prime }(0)\mathbf{\text{tidak terdefinisi}}\\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

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  $h={\displaystyle \frac{1}{2}g{t}^2}$  dengan  $h$  adalah tinggi benda dengan percepatan grafitasinya adalah  $g=10m/s^2$ dan  $t$  adalah waktu tempuh.

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

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

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

$△v={\displaystyle \frac{△y}{△t}={\displaystyle \frac{y_n-y_1}{t_n-t_1}}}$

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

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

Coba perhatikanlah tabel berikut:

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

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:

$\begin{array}{|cc|}\hline \text{Laju perubahan rata-rata}& \text{Laju perubahan sesaat}\\ \begin{array}{rl}& \\ & {\displaystyle \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}={\displaystyle \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}}}\\ & \end{array}& \begin{array}{rl}& \\ & \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(a+h)-f(a)}{h}}\\ & \end{array}\\ \hline\end{array}$.

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

$\begin{array}{rl}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5(t+h{)}^2-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5(t^2+2th+h^2)-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5t^2+10th+5h^2-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{10th+5h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}10t+5h\\ & =10t\end{array}$

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

Secara matematis, perubahan laju terhadap suatu fungsi di  $x=a$ selanjutnya dinotasikan dengan $f^{\prime }(x)$ dan didefiniskan dengan:

$\boxed{{\displaystyle f^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}}}$

Bentuk di atas dinamakan dengan derivatif atau turunan pertama pada fungsi  $f(x)$  dan dinotasikan dengan  $f^{\prime }(x)$ dan proses pencarian derivatif ini dinamakan differensial.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Jika}g(x)=3x-5,\text{hitunglah laju perubahan }\\ & \text{fungsi}g\text{di}x=2\\ & \\ & \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa}g(x)=3x-5\\ & \begin{array}{|cc|}\hline \text{Cara Pertama}& \text{Cara Kedua}\\ \begin{array}{rl}g(x)& =3x-5\\ g(2)& =3.2-5=1\\ g^{\prime }(2)& =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{g(x)-g(2)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{(3x-5)-(1)}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{3x-6}{x-2}}\\ & =\underset{x\to 2}{\text{Lim}}{\displaystyle 3}\\ & =3\end{array}& \begin{array}{rl}g(2)& =1\\ g(2+& h)=3(2+h)-5=3h+1\\ g(2+& h)-g(2)=3h\\ g^{\prime }(2)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{g(2+h)-g(2)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{3h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}3\\ & =3\\ & \end{array}\\ \hline\end{array}\\ & \text{Jadi, laju perubahan fungsi}g\text{di}x=2\text{adalah}3\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui}f(x)=2022x^2,\text{tentukanlah}{f}^{\prime }(x)\text{dan}{f}^{\prime }(1)\\ & \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline f^{\prime }(x)& f^{\prime }(1)\\ \begin{array}{rl}f(x)& =2022x^2\\ f(x+h)& =2022(x+h{)}^2\\ & =2022(x^2+2xh+h^2)\\ & =2022x^2+4044xh+2022h^2\\ f^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{(2022x^2+4044xh+2022h^2)-\left(2022x^2\right)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4044xh+2022h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 4044x+2022h}\\ & =4044x\end{array}& \begin{array}{rl}{f}^{\prime }(x)& =4044x\\ \text{maka},& \\ f^{\prime }(1)& =4044.1\\ & =4044\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah kecepatan jika diketahui}f(t)=\sin t\\ & \text{saat}t\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{f}^{\prime }(t)=v(t)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (t+h)-\sin t}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2t+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2t+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2t+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2t+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2t)}\\ & =\cos t\end{array}\end{array}$

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

DAFTAR PUSTAKA

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