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: 10


$f'(x)=\frac{-9 (5x^4-1)}{(x^5-x+1)^{10}} $

Work Step by Step

Given that $f(x)=(x^5-x+1)^{-9} $ We can first write it as $ f ( x ) =(x^5-x+1)^{-9} $ so that it is easier to deal with. Next we write this as$ f(x)=(g(x))^{-9}$, where $f(x) $ is the outside function and $ g(x)=x^5-x+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^5-x+1)^{-9} $ $ f (x)=(g (x))^{-9}$ $f'(x)=-9(g ( x ))^{-9-1}×(g' ( x )) $ $f'(x)=-9(g ( x ))^{-10}×(5x^{5-1}-1$ $f'(x)=-9(x^5-x+1)^{-10}×(5x^{4}-1) $ $f'(x)=-9(5x^{4}-1)(x^5-x+1)^{-10}$ $f'(x)=\frac{-9 (5x^4-1)}{(x^5-x+1)^{10}}$
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