Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2* The Natural Logarithmic Functions - 6.2* Exercises: 64

Answer

$y'=\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}[\frac{12x^{2}}{(x^{3}+1)}+2cotx-\frac{1}{3x}]$

Work Step by Step

Use logarithmic differentiation to find the derivative of the function. $y=\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}$ Taking logarthimic on both sides . $lny=ln[\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}]$ Use logarithmic properties $ln(xy)=lnx+lny$ , $ln(\frac{x} {y})=lnx-lny$ and $ln(x^{y})=ylnx$. $lny=4ln{(x^{3}+1)}+2lnsinx-\frac{1}{3}lnx$ Differentiate with respect to $x$ $y'=y[\frac{12x^{2}}{(x^{3}+1)}+2cotx-\frac{1}{3x}$ Hence, $y'=\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}[\frac{12x^{2}}{(x^{3}+1)}+2cotx-\frac{1}{3x}]$
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