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 1176: 79


See the explanation below.

Work Step by Step

Consider the function $f\left( x \right)={{x}^{n}}$ , Find the derivative of the function $f\left( x \right)={{x}^{n}}$ using the formula ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)-f\left( x \right)}{h}$ , ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$ Now, expand ${{\left( x+h \right)}^{n}}$ by the binomial formula, ${{\left( x+h \right)}^{n}}={{x}^{n}}+n{{x}^{n-1}}h+n\left( n-1 \right){{x}^{n-2}}{{h}^{2}}+\ldots +nx{{h}^{n-1}}+{{h}^{n}}$ Now, substitute the value of ${{\left( x+h \right)}^{n}}$ in ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$ and solve further, $\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{x}^{n}}+n{{x}^{n-1}}h+n\left( n-1 \right){{x}^{n-2}}{{h}^{2}}+\ldots +nx{{h}^{n-1}}+{{h}^{n}}-{{x}^{n}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{n{{x}^{n-1}}h+n\left( n-1 \right){{x}^{n-2}}{{h}^{2}}+\ldots +nx{{h}^{n-1}}+{{h}^{n}}}{h} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( n{{x}^{n-1}}+n\left( n-1 \right){{x}^{n-2}}h+\ldots +nx{{h}^{n-2}}+{{h}^{n-1}} \right) \end{align}$ Now, take the limit inside and solve further, $\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,n{{x}^{n-1}}+\underset{h\to 0}{\mathop{\lim }}\,n\left( n-1 \right){{x}^{n-2}}h+\ldots +\underset{h\to 0}{\mathop{\lim }}\,nx{{h}^{n-2}}+\underset{h\to 0}{\mathop{\lim }}\,{{h}^{n-1}} \\ & =n{{x}^{n-1}} \end{align}$ Thus, ${f}'\left( x \right)=n{{x}^{n-1}}$
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