Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.6 Trigonometric Functions - Exercises - Page 140: 32


$$y= \frac{2-\pi}{4}x+\frac{\pi^2}{16}$$

Work Step by Step

Given $$y=x \cos ^{2} x, \quad x=\frac{\pi}{4}$$ Since $y(\pi/4)= \pi/8$ and \begin{align*} y'&= \frac{d}{dx}\left(x\right)\cos ^2\left(x\right)+\frac{d}{dx}\left(\cos ^2\left(x\right)\right)x\\ &=\cos ^2\left(x\right)-x\sin \left(2x\right)\\ y'\bigg|_{x=\pi/4}&= \frac{2-\pi}{4} \end{align*} Then the tangent line equation is given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y- \pi/8}{x-\pi/4} &= \frac{2-\pi}{4}\\ y- \pi/8&= \frac{2-\pi}{4}[x-\pi/4]\\ y&= \frac{2-\pi}{4}x+\frac{\pi^2}{16} \end{align*}
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