Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 1

Answer

$$\frac{dy}{dx}=\frac{4}{3\sqrt[3]{(1+4x)^2}}$$

Work Step by Step

$$y=\sqrt[3]{1+4x}$$ $$\frac{dy}{dx}=\frac{d(\sqrt[3]{1+4x})}{dx}$$ $$\frac{dy}{dx}=\frac{d[(1+4x)^{1/3}]}{dx}$$ Let $u=1+4x$ and $y=u^{1/3}$. Then according to Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{d(u^{1/3})}{du}\frac{d(1+4x)}{dx}$$ $$\frac{dy}{dx}=\frac{1}{3}u^{-2/3}\times(0+4\times1)$$ $$\frac{dy}{dx}=\frac{1}{3\sqrt[3]{u^2}}\times4$$ $$\frac{dy}{dx}=\frac{4}{3\sqrt[3]{(1+4x)^2}}$$
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