Find dy/dx y=sin(cos(x))

$y = \sin (\cos (x))$

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\sin (\cos (x)))$

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Differentiate the right side of the equation.

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Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = \cos (x)$ .

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To apply the Chain Rule, set $u$ as $\cos (x)$ .

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [\cos (x)]$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [\cos (x)]$

Replace all occurrences of $u$ with $\cos (x)$ .

$\cos (\cos (x)) \frac{d}{d x} [\cos (x)]$

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$\cos (\cos (x)) (- \sin (x))$

Simplify the expression.

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Move $- 1$ to the left of $\cos (\cos (x))$ .

$- 1 \cdot \cos (\cos (x)) \sin (x)$

Rewrite $- 1 \cos (\cos (x))$ as $- \cos (\cos (x))$ .

$- \cos (\cos (x)) \sin (x)$

Reform the equation by setting the left side equal to the right side.

$y' = - \cos (\cos (x)) \sin (x)$

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \cos (\cos (x)) \sin (x)$

Find dy/dx y=sin(cos(x))

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