# 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

Definition of a derivative function

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* = (2*x*^{3} - 5)(4*x*^{2} + *x*)*y*' is equal to,

the derivative of 2*x*^{3} - 5, (2⋅3*x*^{2} - 0)

times (4*x*^{2} + *x*)

plus (2*x*^{3} - 5)

times, the derivative of 4*x*^{2} + *x*, (4⋅2*x*^{1} + 1*x*^{0}).

Derivatives of polynomials

Add the *x*^{4} terms. (gray)

And add the *x*^{3} terms. (dark gray)

So *f*'(*x*) = 40*x*^{4} + 8*x*^{3} - 40*x* - 5.

## Example 2

*y* = (*x*^{5} - 2*x*)(7*x*^{2} + 3)*y*' is equal to,

the derivative of *x*^{5} - 2*x*, (5*x*^{4} - 2⋅1*x*^{0})

times (7*x*^{2} + 3)

plus (*x*^{5} - 2*x*)

times, the derivative of 7*x*^{2} + 3, (7⋅2*x*^{1} + 0).

Derivatives of polynomials

So *f*'(*x*) = (5*x*^{4} - 2)(7*x*^{2} + 3) + (*x*^{5} - 2*x*)⋅14*x*.

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.