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 - Page 158: 52



Work Step by Step

First we will find the derivative: $$y'=\left(\sqrt{1+x^3}\right)'=\frac{1}{2\sqrt{1+x^3}}(1+x^3)'=\frac{3x^2}{2\sqrt{1+x^3}}$$ 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 the point $(x_{0},y_{0})=(2,3)$ we get: $$y-3=\left.\frac{3x^2}{2\sqrt{1+x^3}}\right|_{x_{0}=2}(x-2)\Rightarrow y=\frac{3\cdot2^2}{2\sqrt{1+2^3}}(x-2)+3\Rightarrow y=\frac{12}{2\cdot3}(x-2)+3\Rightarrow y=2x-1$$
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