Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.6 - Inverse Trigonometric Functions - Exercises 7.6 - Page 423: 119

Answer

$f(x)=g(x)-\displaystyle \frac{\pi}{2},\quad x\geq 0$

Work Step by Step

$\displaystyle \frac{d}{dx}[f(x)]=\frac{d}{dx}[\sin^{-1}u],\quad u=\frac{x-1}{x+1}$ $=\displaystyle \frac{1}{\sqrt{1-u^{2}}}\cdot\frac{du}{dx}$ $=\displaystyle \frac{1}{\sqrt{1-(\frac{x-1}{x+1})^{2}}}\cdot\frac{1(x+1)-(x-1)(1)}{(x+1)^{2}}$ $=\displaystyle \frac{2}{(x+1)^{2}\sqrt{\frac{(x+1)^{2}-(x-1)^{2}}{(x+1)^{2}}}}\qquad $... x is nonnegative, so $\sqrt{(x+1)^{2}}=(x+1)$ $=\displaystyle \frac{2}{(x+1)\sqrt{2x+2x}}$ $=\displaystyle \frac{2}{(x+1)2\sqrt{x}}$ $=\displaystyle \frac{1}{(x+1)\sqrt{x}}$ $\displaystyle \frac{d}{dx}[g(x)]=\frac{d}{dx}[2\tan^{-1}u],\quad u=\sqrt{x}$ $=2\displaystyle \cdot\frac{1}{1+u^{2}}\cdot\frac{du}{dx}$ $=\displaystyle \frac{2}{1+(\sqrt{x})^{2}}\cdot\frac{1}{2}x^{-1/2}$ $=\displaystyle \frac{1}{(x+1)\sqrt{x}}$ The functions have equal derivatives, which means that they have graphs vertically shifted in regard to each other. This means that $f(x)=g(x)+C$ Set x=0 (to find C) $\displaystyle \sin^{-1}(\frac{0-1}{0+1})=2\tan^{-1}\sqrt{1}+C$ $\sin^{-1}(-1)=2\tan^{-1}1+C$ ... $y=\sin^{-1}x$ is the number in $[-\pi/2, \pi/2]$ for which $\sin y=x.$ ... $y=\tan^{-1}x$ is the number in $(-\pi/2, \pi/2)$ for which $\tan y=x.$ $-\displaystyle \frac{\pi}{2}=0+C$ $C=-\displaystyle \frac{\pi}{2}$ So, $f(x)=g(x)-\displaystyle \frac{\pi}{2},\quad x\geq 0$
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