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: 46

Answer

The equation for the tangent line for $x=2$ is $$y=\frac{135}{16}x-\frac{27}{2}$$

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 $x_0=2$ and $f(x)=(x-\frac{1}{x})^3$. $$f'(x)=\Big(\Big(x-\frac{1}{x}\Big)^3\Big)'=3\Big(x-\frac{1}{x}\Big)^2\Big(x-\frac{1}{x}\Big)'=3\Big(x-\frac{1}{x}\Big)^2(1-(-\frac{1}{x^2}))=3\Big(x-\frac{1}{x}\Big)^2\frac{x^2+1}{x^2}$$ Now for $x=2$ we have: $$f(2)=(2-\frac{1}{2})^3=\frac{27}{8}$$ $$f'(2)=3\Big(2-\frac{1}{2}\Big)^2\frac{2^2+1}{2^2}=3\frac{9}{4}\cdot\frac{5}{4}=\frac{135}{16}$$ So the equation for the tangent line at $x=2$ is: $$y-\frac{27}{8}=\frac{135}{16}(x-2)\Rightarrow y=\frac{135}{16}x-\frac{135}{8}+\frac{27}{8}\Rightarrow y=\frac{135}{16}x-\frac{27}{2}$$
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