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: 4

Answer

$f'(x) = -cos(cot(x))csc^{2}(x)$

Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$. Original expression: $y = sin(cot(x))$ $u = g(x) = cot(x)$ $y = f(u) = sin(u)$ 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}(cot(x)) $ Derive: $f'(u) = cos(u)\times-csc^2(x)$ $f'(x) = cos(cot(x))\times-csc^2(x)=-cos(cot(x))csc^{2}(x)$
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