We will proof the product rule from the definition of the derivative. Because we have a function called #h(x)#, we will not use #h# for the difference, but we will use #\epsilon#. \[f'(x)= \lim_{\epsilon \to 0} \dfrac{f(x+\epsilon)-f(x)}{\epsilon}= \lim_{\epsilon \to 0} \dfrac{g(x+\epsilon)\cdot h(x+\epsilon)-g(x)\cdot h(x)}{\epsilon}\] We now add and subtract #g(x)\cdot h(x+\epsilon)# to the numerator, this won't make a difference in value. \[f'(x)= \lim_{\epsilon \to 0}\dfrac{g(x+\epsilon)\cdot h(x+\epsilon)+g(x)\cdot h(x+\epsilon)-g(x)\cdot h(x+\epsilon)-g(x)\cdot h(x)}{\epsilon}\] We split this into two fractions: \[f'(x)= \lim_{\epsilon \to 0} \left( \dfrac{g(x+\epsilon)\cdot h(x+\epsilon)-g(x)\cdot h(x+\epsilon)}{\epsilon}+\dfrac{g(x)\cdot h(x+\epsilon)-g(x)\cdot h(x)}{\epsilon} \right) \] We can bring out #h(x+\epsilon)# in the first fraction and #g(x)# in the second. This gives \[h(x+\epsilon)\cdot\dfrac{g(x+\epsilon)-g(x)}{\epsilon}+g(x)\cdot\dfrac{h(x+\epsilon)-h(x)}{\epsilon}\] Since #h# is differentiable, and thus continuous, we have that #\lim_{\epsilon \to 0} h(x+\epsilon) = h(x)#. By letting #\epsilon# go to #0# we get the definitions of the derivatives of #g(x)# and #h(x)#, and thus we get \[f'(x)=g'(x)\cdot h(x)+g(x)\cdot h'(x)\]