Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.5 The Chain Rule - 2.5 Exercises: 51

Answer

$$y=18x+1$$

Work Step by Step

First we will find the derivative: $$y'=((3x-1)^{-6})'=-6(3x-1)^{-7}(3x-1)'=-6(3x-1)^{-7}\cdot3=-18(3x-1)^{-7}$$ The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula: $$y-y_{0}=y'(x_{0})(x-x_{0})$$ For $(x_{0},y_{0})=(0,1)$ we get: $$y-1=\left.-18(3x_{0}-1)^{-7}\right|_{x_{0}=0}(x-0)\Rightarrow y=-18(3\cdot0-1)^{-7}\cdot x+1\Rightarrow y=-18\cdot(-1)x+1\Rightarrow y=18x+1$$
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