We can prove the above Product Rule by using the definition of the derivative:
#f'(x)=lim_(h->0)(f(x+h)-f(x))/h#
#(fg)'=lim_(h->0)(f(x+h)g(x+h)-f(x)g(x))/h#
Manipulate things around a bit by subtracting out and adding in #f(x+h)g(x)# to the numerator. This is essentially the same as adding #0:#
#(fg)'=lim_(h->0)(f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x))/h#
Break up the limit into two pieces:
#(fg)'=lim_(h->0)(f(x+h)g(x+h)-f(x+h)g(x))/h + lim_(h->0)(f(x+h)g(x)-f(x)g(x))/h#
Factor #f(x+h)# from the numerator of the first limit and #g(x)# from the numerator of the second limit:
#(fg)'=lim_(h->0)(f(x+h)(g(x+h)-g(x)))/h + lim_(h->0)(g(x)(f(x+h)-f(x)))/h#
#(fg)'=(lim_(h->0)f(x+h) * lim_(h->0)(g(x+h)-g(x))/h) + (lim_(h->0)g(x) * lim_(h->0)(f(x+h)-f(x))/h#
The individual limits are:
#lim_(h->0)g(x)=g(x)#
#lim_(h->0)f(x+h)=f(x)#
#lim_(h->0)(f(x+h)-f(x))/h=f'(x)#
#lim_(h->0)(g(x+h)-g(x))/h=g'(x)#
Plugging #f(x), g(x), f'(x), g'(x)# into #(fg)'=(lim_(h->0)f(x+h) * lim_(h->0)(g(x+h)-g(x))/h) + (lim_(h->0)g(x) * lim_(h->0)(f(x+h)-f(x))/h# gives:
#(fg)'=fg'+gf'#
