Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.5 Derivatives of Logarithmic Functions - 4.5 Exercises - Page 241: 56

Answer

$=(x^2+1)^{5x}[\frac{10x^2}{x^2+1}+5\ln (x^2+1)]$

Work Step by Step

We are given $$h(x)=(x^2+1)^{5x}$$ From Exercise 53 we have the idea: $\frac{d}{dx}h(x)=u(x)^{v(x)}[\frac{v(x)u'(x)}{u(x)}+(\ln u(x)v'(x))]$ so: $\frac{d}{dx}h(x)=(x^2+1)^{5x}[\frac{5x(2x)}{x^2+1}+5\ln (x^2+1)]$ $=(x^2+1)^{5x}[\frac{10x^2}{x^2+1}+5\ln (x^2+1)]$
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