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.
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