Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.3 - Derivatives of Trigonometric Functions - 3.3 Exercises: 24

Answer

The equation of the tangent line $(l)$ to $y$ at point $(\pi,\pi)$ would be $$(l): y=2x-\pi$$

Work Step by Step

$$y=x+\tan x$$ 1) Find the derivative of $y$ $$y'=(x+\tan x)'$$ $$y'=1+\sec^2 x$$ 2) Find $y'(\pi)$ $$y'(\pi)=1+\sec^2\pi$$ $$y'(\pi)=1+(-1)^2=2$$ 3) We know $y'(\pi)$, therefore the slope of the tangent line $(l)$ to $y$ at point $A(\pi,\pi)$ is also known, since it equals $y'(\pi)$. So the equation of the tangent line $(l)$ to $y$ at point $A(\pi,\pi)$ would be $$(l): y-(\pi)=y'(\pi)(x-\pi)$$ $$(l): y-\pi=2\times(x-\pi)$$ $$(l): y-\pi=2x-2\pi$$ $$(l): y=2x-\pi$$
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