Mathematics/Lessons/Differentiation/The Product Rule
Pure · Differentiation · Year 2

The Product Rule

When two functions of x are multiplied together, each contributes its own rate of change — the product rule adds the two contributions.

7 min read 3 worked examples

The product rule

When $y=uv$ is a product of two functions of $x$, the product rule states

$\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$

In words: differentiate the first factor and multiply by the second, then add the first factor multiplied by the derivative of the second. The rule is symmetric in $u$ and $v$ — it doesn't matter which factor is labelled which.

Why you can't just multiply the derivatives

$\dfrac{d}{dx}(uv)\ne\dfrac{du}{dx}\times\dfrac{dv}{dx}$ — differentiating each factor separately and multiplying the results gives the wrong answer. The product rule's two-term structure exists precisely because both factors are changing simultaneously, and each contributes to the overall rate of change.

Combining with the chain rule

One or both factors of a product are often themselves composite functions, needing the chain rule to differentiate. Differentiate each factor with whatever method it needs — the power rule, a standard result, or the chain rule — before substituting into the product rule.

Worked example 1

A product needing the chain rule
Differentiate $y=x^3(2x-1)^4$.
MethodApply the product rule, using the chain rule for one factor
1
Let $u=x^3$ and $v=(2x-1)^4$.
Split the product into its two factors.
2
$\dfrac{du}{dx}=3x^2$. For $v$, the chain rule gives $\dfrac{dv}{dx}=4(2x-1)^3\times2=8(2x-1)^3$.
Differentiate each factor; $v$ needs the chain rule since it's composite.
3
$\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}=x^3\times8(2x-1)^3+(2x-1)^4\times3x^2$.
Substitute into the product rule.
4
$\dfrac{dy}{dx}=8x^3(2x-1)^3+3x^2(2x-1)^4$.
Write out the result.
5
Factorising: $\dfrac{dy}{dx}=x^2(2x-1)^3\big[8x+3(2x-1)\big]=x^2(2x-1)^3(14x-3)$.
Take out the common factor $x^2(2x-1)^3$ and simplify — the tidiest final form.
▪ solved

Worked example 2

A product with a standard trig derivative
Differentiate $y=2x^4\sin x$.
MethodApply the product rule
1
Let $u=2x^4$ and $v=\sin x$, so $\dfrac{du}{dx}=8x^3$ and $\dfrac{dv}{dx}=\cos x$.
Split the product and differentiate each factor.
2
$\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}=2x^4\cos x+8x^3\sin x$.
Substitute into the product rule.
3
$\dfrac{dy}{dx}=2x^3(x\cos x+4\sin x)$.
Factor out $2x^3$ for a tidier final form.
▪ solved

Worked example 3

Finding stationary points of a product · Challenge
Find the values of $x$ in $0\le x\le2\pi$ for which $y=e^x\cos x$ has a stationary point.
MethodDifferentiate, factorise, then solve
1
Let $u=e^x$ and $v=\cos x$, so $\dfrac{du}{dx}=e^x$ and $\dfrac{dv}{dx}=-\sin x$.
Split the product and differentiate each factor.
2
$\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}=e^x(-\sin x)+\cos x\,e^x=e^x(\cos x-\sin x)$.
Substitute into the product rule, then factor out $e^x$.
3
Since $e^x>0$ for every $x$, stationary points occur exactly where $\cos x-\sin x=0$, i.e. $\tan x=1$.
A product of a never-zero factor and another factor is zero only when the second factor is zero.
4
In $0\le x\le2\pi$: $x=\dfrac{\pi}{4}$ or $x=\dfrac{5\pi}{4}$.
$\tan x=1$ has period $\pi$.
▪ solved
Look deeper

Factoring out the never-zero exponential term is a recurring trick: whenever a product involves $e^x$ (or any exponential), it can never contribute a zero itself, so it's always safe to divide it out when solving $\dfrac{dy}{dx}=0$.

Common mistake

Multiplying the two derivatives together instead of using the two-term product rule structure, and forgetting the chain rule on a composite factor — e.g. differentiating $(2x-1)^4$ inside a product as if it were just $2x-1$.

Check yourself

Differentiate $y=x^2e^{3x}$.

Method
1
$u=x^2$, $v=e^{3x}$. $\dfrac{du}{dx}=2x$, and by the chain rule $\dfrac{dv}{dx}=3e^{3x}$.
Differentiate each factor.
2
$\dfrac{dy}{dx}=x^2(3e^{3x})+e^{3x}(2x)=3x^2e^{3x}+2xe^{3x}$.
Substitute into the product rule.
3
$\dfrac{dy}{dx}=xe^{3x}(3x+2)$.
Factor out $xe^{3x}$ for the tidiest final form.
▪ solved
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The Chain Rule