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: 94


$$y'=(x\cos x\ln x+\sin x)x^{\sin x-1}$$

Work Step by Step

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