# Chain rule for differentiation

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## Statement for two functions

Suppose $f$ and $g$ are functions such that $g$ is differentiable at a point $x = x_0$, and $f$ is differentiable at $g(x_0)$. Then the composite $f \circ g$ is differentiable at $x_0$, and we have:

$\frac{d}{dx}[f(g(x))]|_{x = x_0} = f'(g(x_0))g'(x_0)$

In terms of general expressions:

$\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$

In point-free notation, we have:

$(f \circ g)' = (f' \circ g) \cdot g'$

where $\cdot$ denotes the pointwise product of functions.