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 - Page 158: 9


$f'(x)=-2(x^3-\frac{7}{x})^{-3}(3x^2+\frac{7}{x^2}) $

Work Step by Step

Given that $f(x)=(x^3-\frac{7}{x})^{-2}$ We can write this as $f(x)=(g(x))^{-2}$, where $f(x)$ is the outside function and $g(x)=x^3-\frac{7}{x} = x^3-7x^{-1}$ is the inside function. To find the answer we need to differentiate the outside function $f(x) $and times it by the derivative of the inside function $g(x)$. $f(x)=(x^3-\frac{7}{x})^{-2}$ $ f (x)=(g (x))^{-2}$ $f'(x)=-2(g ( x ))^{-2-1}×(g' ( x )) $ $f'(x)=-2(g ( x ))^{-3}×(3x^{3-1} +7x^{-1-1})$ $f'(x)=-2(x^3-\frac{7}{x} )^{-3}×(3x^{2} +7x^{-2}) $ $f'(x)=-2(x^3-\frac{7}{x})^{-3}(3x^2+\frac{7}{x^2}) $
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