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: 44

Answer

$$\frac{dy}{dx}=\ln2\ln3\ln4\times2^{3^{4^x}}\times3^{4^x}\times4^x$$

Work Step by Step

$$y=2^{3^{4^x}}$$ $$\frac{dy}{dx}=\frac{d(2^{3^{4^x}})}{dx}$$ Let $u=4^x$, then $y=2^{3^u}$. According to Chain Rule, we have $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{{d(2^{3^{u}})}}{du}\frac{d(4^x)}{dx}$$ Let $v=3^u$, then $y=2^v$. According to Chain Rule, we have $$\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{du}\frac{du}{dx}$$ $$\frac{dy}{dx}=\frac{{d(2^{v})}}{dv}\frac{d(3^u)}{du}\frac{d(4^x)}{dx}$$ $$\frac{dy}{dx}=\ln2\times2^v\times\ln3\times3^u\times\ln4\times4^x$$ $$\frac{dy}{dx}=\ln2\ln3\ln4\times2^{3^{4^x}}\times3^{4^x}\times4^x$$
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