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

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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)