Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 45

Answer

The equation for the tangent line at $x=-\frac{\pi}{2}$ is $y=-1.$

Work Step by Step

The equation for the tangent line at $x_0$ is: $$y-f(x_0)=f'(x_0)(x-x_0)$$ Here is $x_0=-\frac{\pi}{2}$ and $f(x)=\sec^3(\frac{\pi}{2}-x)$. $$f'(x)=(\sec^3(\frac{\pi}{2}-x))'=3\sec^2(\frac{\pi}{2}-x)(\sec(\frac{\pi}{2}-x))'= 3\sec^2(\frac{\pi}{2}-x)\sec(\frac{\pi}{2}-x)\tan(\frac{\pi}{2}-x)(\frac{\pi}{2}-x)' 3\sec^3(\frac{\pi}{2}-x)\tan(\frac{\pi}{2}-x)\cdot(-1)=-3\sec^3(\frac{\pi}{2}-x)\tan(\frac{\pi}{2}-x)$$ Now for $x=-\frac{\pi}{2}$ we have: $$f(-\frac{\pi}{2})=\sec^2(\frac{\pi}{2}+\frac{\pi}{2})=\sec^3\pi=-1$$ $$f'(-\frac{\pi}{2})=-3\sec^3(\frac{\pi}{2}+\frac{\pi}{2})\tan(\frac{\pi}{2}+\frac{\pi}{2})=-3\sec^3\pi\tan\pi=0$$ So the equation for the tangent line at $x=-\frac{\pi}{2}$ is: $$y-(-1)=0\cdot(x-(-\frac{\pi}{2}))\Rightarrow y=-1$$
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