Product Rule – Sam Artigliere's Blog

Statement

$\frac{d}{dx}\left[ f(x)g(x) \right]=f(x)g^\prime(x) + g(x)f^\prime(x)$

Proof

By the definition of the derivative:

$\frac{d}{dx}\left[ f(x)g(x) \right]&=&\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x+h) - f(x)g(x)}{h}$

$-f(x+h)g(x) + f(x+h)g(x)=0$ . Add this to the numerator of the equation above. Since it equals zero, it won’t change the value of the numerator:

$\frac{d}{dx}\left[ f(x)g(x) \right]&=&\displaystyle\lim_{h\to 0}\frac{f(x+h)g(x+h) -f(x+h)g(x) + f(x+h)g(x)- f(x)g(x)}{h}$

Rearrange:

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

This is equivalent to:

$\frac{d}{dx}\left[ f(x)g(x)\right] = \underbrace{\left( \displaystyle\lim_{h\to 0}f(x+h)\right)}_{f(x)}\underbrace{ \left(\displaystyle\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\right)}_{g^\prime(x)} + \underbrace{\left(\displaystyle\lim_{h\to 0}g(x)\right)}_{g(x)}\underbrace{\left(\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \right)}_{f^\prime(x)}$

So,

$\frac{d}{dx}\left[ f(x)g(x)\right] =f(x)g^\prime(x) + g(x)f^\prime(x)$

which is what we were trying to prove.