Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.7 The Chain Rule - 3.7 Exercises - Page 191: 23


$y'= \frac{-315x^2}{(7x^3+1)^4}$

Work Step by Step

$\frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x)$ $y=5(7x^3+1)^{-3}$ Because 5 is a constant, the answer will be $5 \times \frac{d}{dx}[(7x^3+1)^{-3}]$ $\frac{d}{dx}[(7x^3+1)^{-3}] = (-3)(7x^3+1)^{-4} \times 21x^2$ $5 \times \frac{d}{dx}[(7x^3+1)^{-3}] = (5) (-3)(7x^3+1)^{-4} \times 21x^2 = \frac{-315x^2}{(7x^3+1)^4}$
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