Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.8 Implicit Differentiation - Exercises - Page 152: 16


$$s^{\prime}(x)=\frac{-s^{2}\left(2 \sqrt{x+s}+x^{2}\right)}{x^{2}\left(s^{2}+2 \sqrt{x+s}\right)} $$

Work Step by Step

Given $$\sqrt{x+s}=\frac{1}{x}+\frac{1}{s}$$ Differentiate both sides with respect to $x$, \begin{align*} \frac{d}{dx}\sqrt{x+s}&= \frac{d}{dx}\frac{1}{x}+ \frac{d}{dx}\frac{1}{s}\\ \frac{1+s'(x)}{2\sqrt{x+s} }&= \frac{-1}{x^2}-\frac{s'(x)}{s^2}\\ \frac{1 }{2\sqrt{x+s} }+\frac{s'(x)}{2\sqrt{x+s} }&= \frac{-1}{x^2}-\frac{s'(x)}{s^2}\\ \frac{1}{2 \sqrt{x+s}} s^{\prime}(x)+\frac{1}{s^{2}} s^{\prime}(x)&=\frac{-1}{x^{2}}-\frac{1}{2 \sqrt{x+s}}\\ \frac{s^{2}+2 \sqrt{x+s}}{2 s^{2} \sqrt{x+8}} s^{\prime}(x)&=\frac{-2 \sqrt{x+s}-x^{2}}{2 x^{2} \sqrt{x+s}} \end{align*} Then $$s^{\prime}(x)=\frac{-s^{2}\left(2 \sqrt{x+s}+x^{2}\right)}{x^{2}\left(s^{2}+2 \sqrt{x+s}\right)} $$
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