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: 15

Answer

$f'(x)=-\frac{2\cos\left(\frac{1}{x^2}\right)}{x^3}$

Work Step by Step

$f(x)=sin(\frac{1}{x^2})$ $f(x)=sin(x^{-2})$ $f'(x)=cos(x^{-2})\times{-2x^{-2-1}}$ $f'(x)=cos(x^{-2})\times{-2x^{-3}}$ $f'(x)=-\frac{2}{x^3}cos(\frac{1}{x^2})$ $f'(x)=-\frac{2\cos\left(\frac{1}{x^2}\right)}{x^3}$
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