proof of product rule

We begin with two differentiable functions $f(x)$ and $g(x)$ and show that their product is differentiable, 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

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}\left[f(x)g(x)\right]$ $=$ $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)-f(x)g(x)}{h}}$

$=$ $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)+f(x+h)g(x)-f(x+h)g(x)-f(x)g(x)}{h}}$

$=$ $\underset{h\to 0}{lim}\left[f(x+h){\displaystyle \frac{g(x+h)-g(x)}{h}}+g(x){\displaystyle \frac{f(x+h)-f(x)}{h}}\right]$

$=$ $\underset{h\to 0}{lim}\left[f(x+h){\displaystyle \frac{g(x+h)-g(x)}{h}}\right]+\underset{h\to 0}{lim}\left[g(x){\displaystyle \frac{f(x+h)-f(x)}{h}}\right]$

$=$ $f(x)g^{\prime }(x)+f^{\prime }(x)g(x).$

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