Proof of the Product Rule
Key Questions

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)+0f(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)#