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: 5

Answer

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

Work Step by Step

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