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: 66



Work Step by Step

$$\tan y=e^x+\ln x$$ Use implicit differentiation here to find $dy/dx$. Differentiate both sides with respect to $x$: $$\frac{d}{dx}(\tan y)=e^x+\frac{1}{x}$$ Apply the Chain Rule for $(\tan y)$: $$\sec^2y\frac{dy}{dx}=e^x+\frac{1}{x}$$ $$\frac{dy}{dx}=\frac{e^x+\frac{1}{x}}{\sec^2y}=\frac{\frac{xe^x+1}{x}}{\sec^2y}$$ $$\frac{dy}{dx}=\frac{xe^x+1}{x\sec^2y}$$
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