Proof of product rule for differentiation using chain rule for partial differentiation

This article proves the product rule for differentiation in terms of the chain rule for partial differentiation.

Statements

Statement of product rule for differentiation (that we want to prove)

uppose $f$ and $g$ are functions of one variable. Then the following is true wherever the right side expression makes sense (see concept of equality conditional to existence of one side):

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

Statement of chain rule for partial differentiation (that we want to use)

Suppose $f,g$ are both functions of one variable and $h$ is a function of two variables. Suppose $u = f(x), v = g(x), w = h(u,v)$. Then:
$\frac{dw}{dx} = \frac{\partial w}{\partial u}\frac{du}{dx} + \frac{\partial w}{\partial v} \frac{dv}{dx} = h_u(u,v)f'(x) + h_v(u,v)g'(x)$

Proof

Given: Functions $f$ and $g$

To prove: $\! \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ wherever the right side makes sense.

Proof:

Consider the function:

$\! h(u,v) := uv$

Its partial derivatives are:

$\! h_u(u,v) = v, \qquad h_v(u,v) = u$ Define:

$\! u := f(x), \qquad v := g(x), \qquad w := uv = h(u,v) = f(x)g(x)$

By the chain rule for partial differentiation, we have:

$\! \frac{dw}{dx} = \frac{\partial w}{\partial u} \frac{du}{dx} + \frac{\partial w}{\partial v}\frac{dv}{dx}$

The left side is $(fg)'(x)$. The right side becomes:

$\! h_u(u,v)f'(x) + h_v(u,v)g'(x)$

This simplifies to:

$\! vf'(x) + ug'(x)$

Plug back the expressions $u = f(x), v = g(x)$ and get:

$\! f'(x)g(x) + f(x)g'(x)$