Product Rule in Differentiation

How to solve the product rule in differentiation problems: formula, proof, examples, and their solutions.

Formula

[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)

First differentiate f(x): f'(x)g(x).
Then differentiate g(x): f(x)g'(x).

Proof

Write -f(x)g(x + h) and +f(x)g(x + h)
between the numerator's two terms.

Factor the first two terms:
[f(x + h) - f(x)]g(x + h).

And factor the last two terms:
f(x)[g(x + h) - g(x)].

Factoring using the distributive property

Split the denominator h.

Write the denominator h
beneath [f(x + h) - f(x)] and [g(x + h) - g(x)].

The the limit of the left gray part is f'(x).
And the limit of the right gray part is g'(x).

Definition of the derivative function

So [f(x)g(x)]' = f'(x)g(x) + f(x)g'(x).

Example 1

y = (2x3 - 5)(4x2 + x)

y' is equal to,
the derivative of 2x3 - 5, (2⋅3x2 - 0)
times (4x2 + x)
plus (2x3 - 5)
times, the derivative of 4x2 + x, (4⋅2x1 + 1x0).

Derivatives of polynomials

And add the x3 terms. (dark gray)

So f'(x) = 40x4 + 8x3 - 40x - 5.

Example 2

y = (x5 - 2x)(7x2 + 3)

y' is equal to,
the derivative of x5 - 2x, (5x4 - 2⋅1x0)
times (7x2 + 3)
plus (x5 - 2x)
times, the derivative of 7x2 + 3, (7⋅2x1 + 0).

Derivatives of polynomials

So f'(x) = (5x4 - 2)(7x2 + 3) + (x5 - 2x)⋅14x.

You don't need to expand f'(x) further
because your goal is to find f'(1), not f'(x).

Put 1 into the f'(x).

Then f'(1) = 16.