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: 56


$$y= 5\left(\pi^{3}\right) \ln \pi(x-1)+\pi^{3}$$

Work Step by Step

Given $$y=\pi^{5 x-2}, \quad x=1$$ Since at $x=1$, $y= \pi ^3$ and $$f'(x)= \pi^{5 x-2}\ln \pi \frac{d}{dx} (5 x-2 ) = \pi^{5 x-2}(5\ln \pi) $$ Then $ m= f'(x)\bigg|_{x=1}=5\pi^3\ln \pi $. Hence, the tangent line is given by \begin{align*} \frac{y-y_1}{x- x_1}&=m\\ \frac{y-\pi^3 }{x- \pi }&=5\pi^3\ln \pi \\ y-\pi^3&= 5\pi^3\ln \pi (x-\pi )\\ y&= 5\left(\pi^{3}\right) \ln \pi(x-1)+\pi^{3} \end{align*}
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