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

First, recall the the the product #fg# of the functions #f# and #g# is defined as #(fg)(x)=f(x)g(x)#. 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) (f(x+h)g(x+h)+0-f(x)g(x))/(h)#

Wich we can rewrite, taking into account that #f(x+h)g(x)-f(x+h)g(x)=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(x+h)(g(x+h)-g(x))/(h) + lim_(h to 0)g(x)(f(x+h)-f(x))/(h)#

Wich give us the product rule

#(fg)^(prime)(x) = f(x)g^(prime)(x)+g(x)f^(prime)(x),#

since:

#lim_(h to 0) f(x+h) = f(x),#

#lim_(h to 0)(g(x+h)-g(x))/(h) = g^(prime)(x),#

#lim_(h to 0) g(x)=g(x),#

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