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: 23

Answer

The equation of the tangent line $(l)$ to $y$ at point $A(\pi,-1)$ would be $$(l): y=x-\pi-1$$

Work Step by Step

$$y=\cos x-\sin x$$ 1) Find the derivative of $y$ $$y'=(\cos x-\sin x)'$$ $$y'=-\sin x-\cos x$$ 2) Find $y'(\pi)$ $$y'(\pi)=-\sin\pi-\cos\pi$$ $$y'(\pi)=-0-(-1)=1$$ 3) We know $y'(\pi)$, therefore the slope of the tangent line $(l)$ to $y$ at point $A(\pi,-1)$ is also known, since it equals $y'(\pi)$. So the equation of the tangent line $(l)$ to $y$ at point $A(\pi,-1)$ would be $$(l): y-(-1)=y'(\pi)(x-\pi)$$ $$(l): y+1=1\times(x-\pi)$$ $$(l): y=x-\pi-1$$
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