University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 93


$$y'=(\ln \sin x+x\cot x)(\sin x)^x$$

Work Step by Step

$$y=(\sin x)^x$$ We can also use logarithmic differentiation to solve these types of problems. Take the natural logarithm of both sides: $$\ln y=\ln(\sin x)^x$$ $$\ln y =x\ln\sin x$$ Now use implicit differentiation to find $y'$ with respect to $x$: $$(\ln y)' =(x\ln\sin x)'$$ $$\frac{y'}{y}=\ln \sin x+x\times\frac{1}{\sin x}\times(\sin x)'$$ $$\frac{y'}{y}=\ln \sin x+\frac{x\cos x}{\sin x}=\ln \sin x+x\cot x$$ Find $y'$: $$y'=(\ln \sin x+x\cot x)y=(\ln \sin x+x\cot x)(\sin x)^x$$
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