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: 21

Answer

The equation of the tangent line $(l)$ to $y$ at point $(0,1)$ would be $$(l): y=x+1$$

Work Step by Step

$$y=\sin x+\cos x$$ 1) Find the derivative of $y$ $$y'=(\sin x+\cos x)'$$ $$y'=\cos x-\sin x$$ 2) Find $y'(0)$ $$y'(0)=\cos 0-\sin0$$ $$y'(0)=1-0=1$$ 3) We know $y'(0)$, therefore the slope of the tangent line $(l)$ to $y$ at point $A(0,1)$ is also known, since it equals $y'(0)$. So the equation of the tangent line $(l)$ to $y$ at point $A(0,1)$ would be $$(l): y-1=y'(0)(x-0)$$ $$(l): y-1=1\times x$$ $$(l): y=x+1$$
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