# How do you differentiate f(x)=sin(cos(3x))?

Nov 18, 2016

$f ' \left(x\right) = - 3 \sin \left(3 x\right) \cos \left(\cos 3 x\right)$

#### Explanation:

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$f \left(x\right)$ is composed of two functions $\text{ "sinx " "and" } \cos x$
$f \left(x\right)$ is differentiated by applying chain rule .
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Let $u \left(x\right) = \sin x \text{ " and " } v \left(x\right) = \cos 3 x$
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Then, $\text{ } f \left(x\right) = u \circ v \left(x\right)$
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The differentiation of $f \left(x\right)$ is :
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$f ' \left(x\right) = \left(u \circ v \left(x\right)\right) ' = u ' \left(v \left(x\right)\right) \times v ' \left(x\right)$
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Computing $u ' \left(v \left(x\right)\right)$
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$u ' \left(x\right) = \cos x$
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$u ' \left(v \left(x\right)\right) = \cos \left(v \left(x\right)\right) = \cos \left(\cos 3 x\right)$
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Computing $v ' \left(x\right)$
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$v ' \left(x\right) = - 3 \sin 3 x$
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$f ' \left(x\right) = \cos \left(\cos 3 x\right) \times \left(- 3 \sin 3 x\right)$

$f ' \left(x\right) = - 3 \sin \left(3 x\right) \cos \left(\cos 3 x\right)$