The Product Rule: Formula, Proof, and Examples

The product rule in calculus is used to differentiate the product of two or more functions.

For a product of two functions, the product rule is equal to the first function multiplied by the derivative of the second function plus the second function multiplied by the derivative of the first function.

The product rule is useful to learn before learning the related quotient rule.

In this lesson you will learn how to use the product rule including lots of examples and practice questions.

Product Rule Formula

Here is the product rule formula:

For $ y = uv $,

$ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} $

Using function notation, we can also write the product rule like this:

$$ \frac{d}{dx} [f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) $$

A proof of the product rule formula can be found at the bottom of this page.

Product Rule Examples

Example 1:

Differentiate $ y = (x^2 - 6)(5x^3 + 2x) $ with respect to x.

Solution:

Let $ u = x^2 - 6 $ and let $ v = 5x^3 + 2x $

Then $ \frac{du}{dx} = 2x $ and $ \frac{dv}{dx} = 15x^2 + 2 $

Substitute into the product rule:

$ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} $

$ \frac{dy}{dx} = (x^2 - 6)(15x^2 + 2) + 2x(5x^3 + 2x) $

$ \frac{dy}{dx} = 25x^4 - 84x^2 - 12 $

Example 2:

Find $ \frac{dy}{dx} $ for $ y = x(2x - 3)^4 $.

Solution:

Let $ u = x $ and let $ v = (2x - 3)^4 $.

Then $ \frac{du}{dx} = 1 $.

And $ \frac{dv}{dx} = 8(x - 3)^3 $ (using the chain rule).

Substitute into the product rule:

$$ \frac{dy}{dx} = x \cdot 8(x - 3)^3 + 1 \cdot (x - 3)^4 $$

$$ = 8x(x - 3)^3 + (x - 3)^4 $$

$$ = 8x(x - 3)^3 + (x - 3)(x - 3)^3 $$

$$ = (9x - 3)(x - 3)^3 $$

Note: factorising the derivative in the last step is not usually required but is shown here for clarity.

Example 3:

Differentiate $ \sqrt{2x + 5}(3x - 1) $

Solution:

Let $ u = \sqrt{2x + 5} $ and let $ v = 3x - 1 $

Then $ \frac{du}{dx} = (2x + 5)^{-1/2} $

and $ \frac{dv}{dx} = 3 $

Substitute into the product rule:

$$ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} $$

$$ \frac{dy}{dx} = 3\sqrt{2x + 5} + (3x - 1)(2x + 5)^{-1/2} $$

$$ = \frac{3(2x + 5) + 3x - 1}{\sqrt{2x + 5}} $$

$$ = \frac{9x + 14}{\sqrt{2x + 5}} $$

Product Rule Questions in Context

The product rule can be used in problems such as finding the equations of tangents and normals to a curve and in identifying stationary points and their nature.

Example

Find the gradient of the tangent to the curve $ y = (2 - x)^3(x + 1)^4 $ where $ x = 1 $.

Solution:

Let $ u = (2 - x)^3 $ and let $ v = (x + 1)^4 $.

Then $ \frac{du}{dx} = -3(2 - x)^2 $ and $ \frac{dv}{dx} = 4(x + 1)^3 $.

$ \frac{dy}{dx} = 4(x + 1)^3(2 - x)^3 - 3(2 - x)^2(x + 1)^4 $

Find the gradient of the tangent:

When $ x = 1 $, $ \frac{dy}{dx} = -16 $, so $ m = -16 $.

Find the y-intercept of the tangent:

When $ x = 1 $, $ y = 16 $. Therefore, $ c = y - mx = 16 - (-16) = 32 $.

Thus, the equation of the tangent is $ y = -16x + 32 $.

Product Rule Practice Questions

Question 1

Differentiate $ y = x(3x + 2)^5 $ with respect to $ x $

Product Rule Proof

The product rule states that for two differentiable functions $ f(x) $ and $ g(x) $, the derivative of their product is given by: $$ \frac{d}{dx} [f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x) $$ We begin by using the definition of the derivative: $$ \frac{d}{dx} [f(x) \cdot g(x)] = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) \cdot g(x+\Delta x) - f(x) \cdot g(x)}{\Delta x} $$ We can add and subtract $ f(x+\Delta x) \cdot g(x) $ to manipulate the expression: $$ = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) \cdot g(x+\Delta x) - f(x+\Delta x) \cdot g(x) + f(x+\Delta x) \cdot g(x) - f(x) \cdot g(x)}{\Delta x} $$ Now, factor out terms: $$ = \lim_{\Delta x \to 0} \left( \frac{f(x+\Delta x) \cdot [g(x+\Delta x) - g(x)]}{\Delta x} + \frac{g(x) \cdot [f(x+\Delta x) - f(x)]}{\Delta x} \right) $$ As $ \Delta x \to 0 $, the first term becomes $ f(x) \cdot g'(x) $, and the second term becomes $ f'(x) \cdot g(x) $. So, we have: $$ \frac{d}{dx} [f(x) \cdot g(x)] = f(x) \cdot g'(x) + f'(x) \cdot g(x) $$ This completes the proof of the product rule.