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: 6


$f'(x) = \frac{cos(\sqrt x)}{2\sqrt x}$

Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$. Original expression: $y =sin( \sqrt x)$ $u = g(x) = \sqrt x$ $y = f(u) = sin(u)$ Rewrite $u$: $\sqrt x$ => $x^{\frac{1}{2}}$ Apply the chain rule to find the derivative: $f'(g(x)) \times g'(x) $ = > $f'(u) \times u'$ => $\frac{dy}{du} sin(u) \times\frac{du}{dx}x^{\frac{1}{2}}$ $f'(u) = cos(u)\times \frac{1}{2}x^{-\frac{1}{2}}$ $f'(u) = cos(u)\times \frac{1}{2\sqrt x}$ $f'(x) = cos(\sqrt x) \times\frac{1}{2\sqrt x}$ $f'(x) = \frac{cos(\sqrt x)}{2\sqrt x}$
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