proof of product rule

We begin with two differentiable functions f(x) and g(x) and show that their product is differentiableMathworldPlanetmath, and that the derivative of the product has the desired form.

By simply calculating, we have for all values of x in the domain of f and g that

ddx[f(x)g(x)] = limh0f(x+h)g(x+h)-f(x)g(x)h
= limh0f(x+h)g(x+h)+f(x+h)g(x)-f(x+h)g(x)-f(x)g(x)h
= limh0[f(x+h)g(x+h)-g(x)h+g(x)f(x+h)-f(x)h]
= limh0[f(x+h)g(x+h)-g(x)h]+limh0[g(x)f(x+h)-f(x)h]
= f(x)g(x)+f(x)g(x).

The key argument here is the next to last line, where we have used the fact that both f and g are differentiable, hence the limit can be distributed across the sum to give the desired equality.

Title proof of product rule
Canonical name ProofOfProductRule
Date of creation 2013-03-22 12:28:00
Last modified on 2013-03-22 12:28:00
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 6
Author mathcam (2727)
Entry type Proof
Classification msc 26A24
Related topic Derivative
Related topic ProductRule