Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1177: 80


See the proof below.

Work Step by Step

Consider the function $f\left( x \right)=\sin x$ , Find the derivative of the function $f\left( x \right)=\sin x$ using the formula ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ , ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \left( x+h \right)-\sin x}{h}$ Now, expand $\sin \left( x+h \right)$ by the formula $\sin \left( a+b \right)=\sin a\cos b+\sin b\cos a$ , $\sin \left( x+h \right)=\sin x\cosh +\sinh \cos x$ Now, substitute the value of $\sin \left( x+h \right)$ in ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \left( x+h \right)-\sin x}{h}$ and solve further, $\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\cosh +\sinh \cos x-\sin x}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\cosh -\sin x+\sinh \cos x}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\left( \cosh -1 \right)+\sinh \cos x}{h} \end{align}$ Now, take the limit inside and solve further, $\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin x\left( \cosh -1 \right)}{h}+\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh \cos x}{h} \\ & =\sin x\underset{h\to 0}{\mathop{\lim }}\,\frac{\left( \cosh -1 \right)}{h}+\cos x\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h} \end{align}$ Substitute the values $\text{ }\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h}=1\text{ and }\underset{h\to 0}{\mathop{\lim }}\,\frac{\cosh -1}{h}=0$ in the above expression and solve further. $\begin{align} & {f}'\left( x \right)=\sin x\left( 0 \right)+\cos x\left( 1 \right) \\ & =\cos x \end{align}$ Thus, $f'\left( x \right)=\cos x$
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