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: 51


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

Work Step by Step

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