Product, quotient and chain rules

The product, quotient and chain rules let us differentiate functions built by multiplying, dividing or composing simpler functions. The difficult part is often not the algebra. It is recognising the structure of the function and choosing the correct rule.

What you should know first

You should be able to:

• differentiate powers, exponentials, logarithms and trigonometric functions;
• use function notation such as $f(x)$ and $f'(x)$;
• simplify algebraic fractions and expressions with indices.

Review differentiating standard functions before continuing if any standard derivative is uncertain.

Choosing the correct rule

Look at the main operation holding the function together.

Structure	Example	Rule
Two variable expressions multiplied	$x^2e^x$	Product rule
One variable expression divided by another	$\dfrac{\sin x}{x}$	Quotient rule
One function placed inside another	$(3x-1)^5$	Chain rule

A function may contain more than one structure. For example,

$$y=x^2\sin(3x)$$

has a product on the outside and a composite function, $\sin(3x)$, inside it. We need the product rule and the chain rule.

Before selecting a rule, simplify if doing so removes unnecessary work. For example,

$$\frac{x^3+2x}{x}=x^2+2, \qquad x\ne 0,$$

so its derivative is simply $2x$ on its domain. The quotient rule would work, but it is needlessly long.

The product rule

If

$$y=u(x)v(x),$$

then

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

Equivalently,

$$\boxed{(uv)'=u'v+uv'}.$$

Differentiate each factor in turn while leaving the other unchanged. Both terms are essential.

Example 1: polynomial multiplied by an exponential

Differentiate

$$y=x^2e^x.$$

Choose

$$u=x^2, \qquad v=e^x.$$

Then

$$\frac{du}{dx}=2x, \qquad \frac{dv}{dx}=e^x.$$

Apply the product rule:

$$\begin{aligned} \frac{dy}{dx} &=u\frac{dv}{dx}+v\frac{du}{dx}\\ &=x^2e^x+e^x(2x)\\ &=e^x(x^2+2x)\\ &=xe^x(x+2). \end{aligned}$$

The first line after substitution is already a correct derivative. Factoring is useful if the result will later be set equal to zero.

Example 2: polynomial multiplied by a trigonometric function

Differentiate

$$y=(2x-1)\cos x.$$

Let $u=2x-1$ and $v=\cos x$. Therefore $u'=2$ and $v'=-\sin x$.

$$\begin{aligned} \frac{dy}{dx} &=(2x-1)(-\sin x)+(\cos x)(2)\\ &=2\cos x-(2x-1)\sin x. \end{aligned}$$

The negative derivative of $\cos x$ is a frequent source of sign errors.

Example 3: three factors

For $y=uvw$, apply the product rule repeatedly:

$$\boxed{(uvw)'=u'vw+uv'w+uvw'}.$$

For example, if

$$y=x^2e^x\sin x,$$

then

$$\frac{dy}{dx}=2xe^x\sin x+x^2e^x\sin x+x^2e^x\cos x.$$

Each term differentiates exactly one factor and leaves the other two unchanged.

The quotient rule

If

$$y=\frac{u(x)}{v(x)}, \qquad v(x)\ne 0,$$

then

$$\boxed{\frac{dy}{dx}=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}}.$$

A useful memory pattern is:

$$\boxed{\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}}.$$

The order in the numerator matters. It is denominator times derivative of numerator, minus numerator times derivative of denominator.

Example 4: rational function

Differentiate

$$y=\frac{x^2+1}{x-2}.$$

Set

$$u=x^2+1, \qquad v=x-2,$$

so $u'=2x$ and $v'=1$. Then

$$\begin{aligned} \frac{dy}{dx} &=\frac{(x-2)(2x)-(x^2+1)(1)}{(x-2)^2}\\ &=\frac{2x^2-4x-x^2-1}{(x-2)^2}\\ &=\frac{x^2-4x-1}{(x-2)^2}. \end{aligned}$$

The derivative is defined only where the original function is defined, so $x\ne 2$.

Example 5: trigonometric quotient

Differentiate

$$y=\frac{\sin x}{x^2}.$$

Take $u=\sin x$ and $v=x^2$. Then $u'=\cos x$ and $v'=2x$.

$$\begin{aligned} \frac{dy}{dx} &=\frac{x^2\cos x-(\sin x)(2x)}{x^4}\\ &=\frac{x\cos x-2\sin x}{x^3}. \end{aligned}$$

An alternative is to write $y=x^{-2}\sin x$ and use the product rule. Both methods give the same result.

The chain rule

A composite function has one function inside another. If

$$y=f(g(x)),$$

then

$$\boxed{\frac{dy}{dx}=f'(g(x))g'(x)}.$$

In words, differentiate the outside function while keeping the inside unchanged, then multiply by the derivative of the inside.

Using an intermediate variable makes the logic visible. If $u=g(x)$ and $y=f(u)$, then

$$\boxed{\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}}.$$

This notation behaves like cancellation, but the chain rule is the mathematical justification.

Example 6: a power of a linear expression

Differentiate

$$y=(3x-4)^7.$$

The outside function is seventh power and the inside function is $3x-4$.

$$\begin{aligned} \frac{dy}{dx} &=7(3x-4)^6\times 3\\ &=21(3x-4)^6. \end{aligned}$$

The factor $3$ is the derivative of the inside. Omitting it is the most common chain rule error.

Example 7: trigonometric composition

Differentiate

$$y=\sin(x^2+1).$$

Keep $x^2+1$ unchanged while differentiating the outside sine:

$$\frac{dy}{dx}=\cos(x^2+1)\times 2x =2x\cos(x^2+1).$$

Example 8: exponential composition

If

$$y=e^{5x^2-3x},$$

then

$$\frac{dy}{dx}=e^{5x^2-3x}(10x-3).$$

The exponential is unchanged because the derivative of $e^u$ with respect to $u$ is $e^u$.

Example 9: logarithmic composition

Differentiate

$$y=\ln(2x^3+5).$$

Since $d(\ln u)/du=1/u$,

$$\frac{dy}{dx}=\frac{1}{2x^3+5}\times 6x^2 =\frac{6x^2}{2x^3+5}.$$

For real values, the original logarithm also requires $2x^3+5>0$.

Example 10: more than one chain

Differentiate

$$y=\left(1+\sin(2x)\right)^4.$$

Work from the outside inward:

$$\begin{aligned} \frac{dy}{dx} &=4\left(1+\sin(2x)\right)^3 \times \frac{d}{dx}\left(1+\sin(2x)\right)\\ &=4\left(1+\sin(2x)\right)^3 \times \cos(2x)\times 2\\ &=8\cos(2x)\left(1+\sin(2x)\right)^3. \end{aligned}$$

There are two inside derivatives: first $\sin(2x)$, then $2x$.

Combining the rules

Identify the outermost structure first. That choice determines the first line of the derivative.

Example 11: product rule with a chain rule

Differentiate

$$y=x^3\cos(2x).$$

The outermost structure is a product. Let $u=x^3$ and $v=\cos(2x)$.

$$u'=3x^2, \qquad v'=-2\sin(2x).$$

Therefore

$$\begin{aligned} \frac{dy}{dx} &=3x^2\cos(2x)+x^3\left(-2\sin(2x)\right)\\ &=3x^2\cos(2x)-2x^3\sin(2x). \end{aligned}$$

Example 12: quotient rule with chain rules

Differentiate

$$y=\frac{e^{2x}}{(x+1)^3}.$$

Use $u=e^{2x}$ and $v=(x+1)^3$. Then

$$u'=2e^{2x}, \qquad v'=3(x+1)^2.$$

Hence

$$\begin{aligned} \frac{dy}{dx} &=\frac{(x+1)^3(2e^{2x})-e^{2x}\left(3(x+1)^2\right)}{(x+1)^6}\\ &=\frac{e^{2x}(x+1)^2\left(2(x+1)-3\right)}{(x+1)^6}\\ &=\frac{e^{2x}(2x-1)}{(x+1)^4}, \qquad x\ne -1. \end{aligned}$$

Rewriting as $e^{2x}(x+1)^{-3}$ and using the product rule is equally valid and often shorter.

Example 13: chain rule outside a quotient

Differentiate

$$y=\left(\frac{x-1}{x+1}\right)^5.$$

The outermost structure is fifth power. Let

$$u=\frac{x-1}{x+1}.$$

First apply the chain rule:

$$\frac{dy}{dx}=5u^4\frac{du}{dx}.$$

Now use the quotient rule on $u$:

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

Therefore

$$\frac{dy}{dx} =10\left(\frac{x-1}{x+1}\right)^4\frac{1}{(x+1)^2} =\frac{10(x-1)^4}{(x+1)^6}.$$

This example shows why the outermost structure matters. Starting with the quotient rule would treat the whole numerator and denominator as fifth powers and create much more algebra.

Efficient choices and checking

Several forms can represent the same function, so more than one method may be correct.

• Expand $(x+1)(x^2-2x)$ before differentiating if expansion is quick.
• Rewrite $1/(x^2+1)$ as $(x^2+1)^{-1}$ to use the chain rule.
• Rewrite $e^x/x^2$ as $e^xx^{-2}$ if the product rule looks cleaner.
• Keep factors unexpanded when a later question asks for stationary points.

You can check a derivative by:

1. checking that every composite function contributed its inside derivative;
2. checking quotient signs and the squared denominator;
3. estimating the gradient numerically at one convenient value;
4. differentiating an algebraically equivalent form by another method.

For instance,

$$y=\frac{1}{3x+2}=(3x+2)^{-1}$$

gives

$$y'=-(3x+2)^{-2}\times 3=-\frac{3}{(3x+2)^2}.$$

The negative sign also makes sense because $1/(3x+2)$ decreases on each interval of its domain.

Common misconceptions

Multiplying the derivatives in a product

It is false that

$$(uv)'=u'v'.$$

For example, $x\cdot x=x^2$ has derivative $2x$, while $1\cdot1=1$. Use $u'v+uv'$.

Dividing the derivatives in a quotient

It is false that

$$\left(\frac{u}{v}\right)'=\frac{u'}{v'}.$$

Use $(vu'-uv')/v^2$, or rewrite the quotient using a negative power.

Reversing the quotient numerator

The numerator is $vu'-uv'$, not $uv'-vu'$. Reversing it changes the sign of the answer.

Forgetting the inside derivative

The derivative of $(x^2+1)^5$ is

$$5(x^2+1)^4(2x),$$

not merely $5(x^2+1)^4$.

Applying a rule to the wrong structure

$\sin x+e^x$ is a sum, so differentiate term by term. $\sin(e^x)$ is a composition, so use the chain rule. Brackets and function notation reveal the difference.

Expanding too early

Expansion can hide helpful factors and introduce errors. Decide whether expanded or factorised form better serves the next part of the problem.

Understanding checks

Try these before reading the answers.

1. Differentiate $y=(x^2+3)e^x$.
2. Differentiate $y=\dfrac{2x+1}{x^2+4}$.
3. Differentiate $y=(5-2x)^6$.
4. Differentiate $y=\cos(3x^2)$.
5. Differentiate $y=x\ln(x^2+1)$.
6. Differentiate $y=\dfrac{\sin(2x)}{x}$.
7. Differentiate $y=\left(\dfrac{x}{x+2}\right)^3$.

Answers

1

Use the product rule:

$$\frac{dy}{dx}=2xe^x+(x^2+3)e^x =e^x(x^2+2x+3).$$

2

Use the quotient rule:

$$\begin{aligned} \frac{dy}{dx} &=\frac{(x^2+4)(2)-(2x+1)(2x)}{(x^2+4)^2}\\ &=\frac{-2x^2-2x+8}{(x^2+4)^2}. \end{aligned}$$

3

Use the chain rule:

$$\frac{dy}{dx}=6(5-2x)^5(-2)=-12(5-2x)^5.$$

4

Use the chain rule twice:

$$\frac{dy}{dx}=-\sin(3x^2)(6x)=-6x\sin(3x^2).$$

5

Use the product rule outside and the chain rule inside:

$$\frac{dy}{dx}=\ln(x^2+1)+x\left(\frac{2x}{x^2+1}\right) =\ln(x^2+1)+\frac{2x^2}{x^2+1}.$$

6

Use the quotient rule and chain rule:

$$\frac{dy}{dx} =\frac{x\left(2\cos(2x)\right)-\sin(2x)}{x^2}.$$

7

Use the chain rule outside. Since

$$\frac{d}{dx}\left(\frac{x}{x+2}\right) =\frac{(x+2)-x}{(x+2)^2} =\frac{2}{(x+2)^2},$$

we get

$$\frac{dy}{dx} =3\left(\frac{x}{x+2}\right)^2\frac{2}{(x+2)^2} =\frac{6x^2}{(x+2)^4}.$$

What to learn next

These rules are used throughout calculus. Continue to implicit differentiation to differentiate relations in which $y$ is not isolated, then study stationary points and curve sketching to use derivatives for analysing graphs.

You will also need these techniques in tangents and normals, parametric differentiation and connected rates of change.