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: 60


$$y =\frac{1}{2}(x-4)+\ln 16$$

Work Step by Step

Given $$f(x)=\ln \left(x^{2}\right) \quad x=4$$ Since at $x=4$, $f(x) = \ln 16$ and $$f'(x)= \frac{2x}{x^2}= \frac{2}{x} $$ Then $ m= f'(x)\bigg|_{x=4}=\dfrac{1}{2} $ Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x- x_1}&=m\\ \frac{y-\ln 16 }{x- 4}&=\dfrac{1}{2} \\ y&=\frac{1}{2}(x-4)+\ln 16 \end{align*}
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