Proof of the Chain Rule of Differentiation

Assume the functions are differentiable on suitable open intervals, with the composite argument in the domain of the outer function.

Let \({\mathrm{f} \circ \mathrm{g}(x) = \mathrm{f}[\mathrm{g}(x)]}\). Show that \({[\mathrm{f} \circ \mathrm{g}(x)]' = \mathrm{f}'[\mathrm{g}(x)]\mathrm{g}'(x)}\).