*This short article shows one way of deriving the Product Rule for differentiation. Read it carefully and see if you can follow the argument. You may wish to try to recreate the proof for yourself. To practise applying the Product Rule, try our Mathmo App where you can generate plenty of questions and then check your answers.*

Suppose we have differentiable functions $f(x)$ and $g(x)$, and we want to find the derivative of their product $F(x)=f(x)g(x)$.

$$

\begin{align}

F'(x)&=\lim_{h \to 0}\frac{F(x+h)-F(x)}{h} \\

&= \lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h} \\

&= \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)+f(x+h)g(x)-f(x+h)g(x)}{h}\\

&= \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} \\

&= \lim_{h \to 0} f(x+h)\left(\frac{g(x+h)-g(x)}{h}\right) + \lim_{h \to 0}g(x)\left(\frac{f(x+h)-f(x)}{h}\right) \\

&=f(x)g'(x) + g(x)f'(x)

\end{align}

$$

Extension

Using the Product Rule and the Chain Rule, can you derive the Quotient Rule for differentiation?

Basic page