Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 56


$$ y'=\frac{3\sin x}{2\sqrt{4-3\cos x}}.$$

Work Step by Step

Since $ y=\sqrt{4-3\cos x}=(4-3\cos x)^{1/2}$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$ and recalling that $(\cos x)'=-\sin x $, the derivative $ y'$ is given by $$ y'=\frac{1}{2}(4-3\cos x)^{-1/2}(4-3\cos x)' \\=\frac{-3(-\sin x)}{2\sqrt{4-3\cos x}}=\frac{3\sin x}{2\sqrt{4-3\cos x}}.$$
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