Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.6 - Derivatives of Logarithmic Functions - 3.6 Exercises: 4

Answer

$f'(x) = 2cotx$

Work Step by Step

$f(x) = \ln (sin^{2}x)$ Use the chain rule: $f'(x) = (\frac{1}{sin^{2}x})(2sinx)(cosx)$ $f'(x) = (\frac{2sinx}{sin^{2}x})(cosx)$ Cancel out $sinx$: $f'(x) = (\frac{2}{sinx})(cosx)$ Multiply: $f'(x) = (\frac{2cosx}{sinx})$ Given trigonometric function: $\frac{cosx}{sinx} = cotx$ $f'(x) = 2cotx$
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