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: 47

Answer

The equation for the tangent line at $x=\sqrt\pi$ is $$y=8\sqrt\pi x-8\pi$$

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 $x_0=\sqrt\pi$ and $f(x)=\tan(4x^2)$. $$f'(x)=(\tan(4x^2))'=\sec^2(4x^2)(4x^2)'=\sec^2(4x^2)\cdot8x=8x\sec^2(4x^2)$$ For $x=\sqrt\pi$ we have: $$f(\sqrt\pi)=\tan(4(\sqrt\pi^2)=\tan4\pi=0$$ $$f'(\sqrt\pi)=8\sqrt\pi\sec^2(4(\sqrt\pi)^2)=8\sqrt\pi\sec^2(4\pi)=8\sqrt\pi$$ So, the equation for the tangent line at $x=\sqrt\pi$ is: $$y-0=8\sqrt\pi(x-\sqrt\pi)\Rightarrow y=8\sqrt\pi x-8\pi$$
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