Proof of the Product Rule

Key Questions

• All we need to do is use the definition of the derivative alongside a simple algebraic trick.

First, recall the the the product $f g$ of the functions $f$ and $g$ is defined as $\left(f g\right) \left(x\right) = f \left(x\right) g \left(x\right)$. Therefore, it's derivative is

(fg)^(prime)(x) = lim_(h to 0) ((fg)(x+h)-(fg)(x))/(h) = lim_(h to 0) (f(x+h)g(x+h)-f(x)g(x))/(h)

Now, note that the expression above is the same as

${\lim}_{h \to 0} \frac{f \left(x + h\right) g \left(x + h\right) + 0 - f \left(x\right) g \left(x\right)}{h}$

Wich we can rewrite, taking into account that $f \left(x + h\right) g \left(x\right) - f \left(x + h\right) g \left(x\right) = 0$, as:

lim_(h to 0) 1/h [f(x+h)g(x+h)+(f(x+h)g(x)-f(x+h)g(x))-f(x)g(x)] = lim_(h to 0) 1/h(f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)])

Using the property that the limit of a sum is the sum of the limits, we get:

${\lim}_{h \to 0} f \left(x + h\right) \frac{g \left(x + h\right) - g \left(x\right)}{h} + {\lim}_{h \to 0} g \left(x\right) \frac{f \left(x + h\right) - f \left(x\right)}{h}$

Wich give us the product rule

${\left(f g\right)}^{p r i m e} \left(x\right) = f \left(x\right) {g}^{p r i m e} \left(x\right) + g \left(x\right) {f}^{p r i m e} \left(x\right) ,$

since:

${\lim}_{h \to 0} f \left(x + h\right) = f \left(x\right) ,$
${\lim}_{h \to 0} \frac{g \left(x + h\right) - g \left(x\right)}{h} = {g}^{p r i m e} \left(x\right) ,$
${\lim}_{h \to 0} g \left(x\right) = g \left(x\right) ,$
${\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h} = {f}^{p r i m e} \left(x\right)$