MATEMATIKA INDONESIA: Turunan 3

Contoh 4 :
Turunan pertama dari
$\fn_jvn f(x)=\sqrt[3]{x}$
adalah

Jawab :
$\fn_jvn f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} &\frac{f(x+h)-f(x)}{h} \end{matrix}$
$\fn_jvn f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}$

$\fn_jvn f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}\left ( \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} \right )\left ( \frac{\sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}}}{\sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}}} \right )$

$\fn_jvn f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}\left \frac{x+h-x}{h\left ( \sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}} \right )}$

$\fn_jvn f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}\left \frac{h}{h\left ( \sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}} \right )}$

$\fn_jvn f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}\left \frac{1}{\sqrt[3]{(x+h)^{2}}+\sqrt[3]{(x+h)x}+\sqrt[3]{x^{2}}}$

$\fn_jvn f'(x)=\frac{1}{\sqrt[3]{(x+0)^{2}}+\sqrt[3]{(x+0)x}+\sqrt[3]{x^{2}}}$

$\fn_jvn f'(x)=\frac{1}{3\sqrt[3]{x^{2}}}$

contoh 5
Tetukan turunan pertama dari f(x) = sin x

Jawab :
$\fn_jvn f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} &\frac{f(x+h)-f(x)}{h} \end{matrix}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{sin(x+h)-sinx}{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{\sin x\cos h + \cos x \sin h - \sin x}{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{\sin x\cos h - \sin x + \cos x \sin h }{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{\sin x(\cos h - 1) + \cos x \sin h }{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{\sin x(1 - 2\sin^{2} \frac{1}{2}h - 1) + \cos x \sin h }{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{-2\sin x\sin^{2} \frac{1}{2}h + \cos x \sin h }{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{-2\sin x\sin^{2} \frac{1}{2}h }{h}+\frac{\cos x \sin h }{h}$

$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} -2\sin x\frac{\sin^{2} \frac{1}{2}h }{h}+\cos x \frac{\sin h }{h}$

$\fn_jvn f'(x)=-2\sin x. 0 +\cos x . 1$

$\fn_jvn f'(x)=\cos x$

Cara II :
$\fn_jvn f'(x)=\begin{matrix}lim\\h \to \0 \end{matrix} \frac{sin(x+h)-sinx}{h}$
Dengan memakai rumus
$\sin A- \sin B=2\cos\frac{1}{2}(A+B)sin\frac{1}{2}(A-B)$
maka diperoleh
$f'(x) =\begin{matrix} lim\\ h \to \0 \end{matrix} \frac{2cos\frac{1}{2}(2x+h)sin\frac{1}{2}h}{h}$
$f'(x)=\begin{matrix} \begin{matrix} lim\\ h \to \0 \end{matrix} & 2\cos\frac{1}{2}(2x+h)\frac{sin\frac{1}{2}h}{h} \end{matrix}$

$\fn_jvn f'(x)=2\cos\frac{1}{2}(2x+0).\frac{1}{2}$

$\fn_jvn f'(x)=\cos x$

berlanjut ke turunan 4
materi sebelumnya (turunan 2)

MATEMATIKA INDONESIA

Sabtu, 15 Oktober 2011

Turunan 3

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