For example, if #f(x)=x+1# and #g(x)=x^3+1#, then #f\cdot g# is the function with function rule \[\left(f\cdot g\right)(x)=f(x)\cdot g(x) = \left(x+1\right)\cdot \left(x^3+1\right) = x^4+x^3+x+1\tiny.\]

This means that #(f\cdot g)'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)# for all #x#.

In order to prove the rule, we rewrite difference quotient as follows:

\[\begin{array}{rcl} \text{difference quotient}&=&\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=& \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &=& \frac{(f(x+h)-f(x))g(x)+f(x+h)(g(x+h)-g(x))}{h}\\ &=& \frac{f(x+h)-f(x)}{h}g(x)+f(x+h)\frac{g(x+h)-g(x)}{h}\\ \end{array}\]

Now we can calculate the limit of the difference quotient for #h\to 0#: \[\begin{array}{rcl} \left(f\cdot g\right)'(x) &=& \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=&\lim_{h\to 0}\left( \frac{f(x+h)-f(x)}{h}g(x)+f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &=&\lim_{h\to 0}\left(\frac{f(x+h)-f(x)}{h}g(x)\right)+\lim_{h\to 0}\left(f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &=&\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\cdot \lim_{h\to 0} g(x)+\lim_{h\to 0}f(x+h)\cdot \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=& f'(x)\cdot g(x)+f(x)\cdot g'(x)\tiny.\\ \end{array}\]