Product Rule: (f g)′ = f ′ g + f g′

As with above the Power Rule, so the Product Rule can be proved either through using the definition of the derivative or this can be proved by using Logarithmic Differentiation. We will demonstrate both proofs now.

Proof

It is a much faster proof but does assume that you have read and understood Logarithmic Differentiation and the Implicit Differentiation sections. If you have not so this proof will not make much sense to you.

Firstly write call the product y and get the log of both sides and utilize a property of logarithms upon the right side.

y = f(x)  g(x)    

ln(y) = ln (f(x)  g(x)) = ln f(x) + ln g(x)

Subsequently, we take the derivative of both sides and then solve for y′.

y'/y = (f'(x)/f(x)) + (g'(x)/g(x))                         =>                    y' = y (f'(x)/f(x)) + (g'(x)/g(x))

At last, all we require to do is plug into for y and after that multiply this with the parenthesis and we find the Product Rule.

y = f(x) g(x) (f'(x)/f(x)) + (g'(x)/g(x))  =>                    (fg)' = g(x) f'(x) + f(x) g'(x)