Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 44

Answer

The equation for the tangent line at $x=-3$ is $ y=27\cos26(x+3)-\sin26.$

Work Step by Step

The equation for the tangent line at $x_0$ is: $y-f(x_0)=f'(x_0)(x-x_0)$ Here is $x_0=-3$ and $f(x)=\sin(1+x^3)$ $$f'(x)=(\sin(1+x^3))'=\cos(1+x^3)(1+x^3)'=\cos(1+x^3)\cdot3x^2= 3x^2\cos(1+x^3)$$ Now we have: $$f(-3)=\sin(1+(-3)^3)=\sin(1-27)=\sin(-26)=-\sin26$$ $$f'(-3)=3\cdot(-3)^2\cos(1+(-3)^3)=27\cos(1-27)=27\cos(-26)=27\cos26$$ So, the equation for the tangent line at $x=-3$ is: $$y-(-\sin26)=27\cos26(x-(-3))\Rightarrow y=27\cos26(x+3)-\sin26$$
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