Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 62



Work Step by Step

Given $$y=\ln (\sin x), \quad x=\frac{\pi}{4}$$ Since at $x=\pi/4$, $f(x) = \ln (1/\sqrt{2}) $ and $$f'(x)= \frac{\cos x}{\sin x}= \cot x$$ Then $ m= f'(x)\bigg|_{x=\pi/4}=1 $ Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x- x_1}&=m\\ \frac{y-\ln (1/\sqrt{2}) }{x- \pi/4}&=1 \\ y&=\left(x-\frac{\pi}{4}\right)+\ln(1/\sqrt{2}) \end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.