1.10.4 Product, quotient and chain rules (A-level only)

What you'll learn

How to recognise and use the product, quotient and chain rules.
How to combine several differentiation rules in one calculation.
How derivatives of inverse functions are related.
How the chain rule connects different rates of change.

Before you begin

You should already be able to differentiate powers, exponentials and trigonometric functions. For example,

ddx(xn)=nxn−1,ddx(ex)=ex,ddx(sin⁡x)=cos⁡x.\frac{d}{dx}(x^n)=nx^{n-1}, \qquad \frac{d}{dx}(e^x)=e^x, \qquad \frac{d}{dx}(\sin x)=\cos x.

The new challenge is deciding what to do when functions are multiplied, divided or placed inside one another.

Key Idea

Identify the structure first

Before differentiating, ask whether the function is a product, a quotient or a composite function. Its structure determines the rule you need.

The product rule

A product is the result of multiplying two expressions. For example, $x^2\sin x$ is a product of $x^2$ and $sin⁡x\sin x$ .

You cannot usually differentiate the two factors separately and then multiply their derivatives.

Definition

The product rule

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

dydx=udvdx+vdudx.\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.

In words: differentiate one factor at a time, leaving the other factor unchanged, and add the results.

Example

Differentiating a product

Differentiate $y=x^2\sin x$ .

Choose $u=x^2$ and $v=sin⁡xv=\sin x$ , so
$dudx=2x,dvdx=cos⁡x.\frac{du}{dx}=2x, \qquad \frac{dv}{dx}=\cos x.$
Substitute into the product rule:
$dydx=x2cos⁡x+(sin⁡x)(2x).\frac{dy}{dx} = x^2\cos x+(\sin x)(2x).$
Simplify:
$dydx=x2cos⁡x+2xsin⁡x.\boxed{\frac{dy}{dx}=x^2\cos x+2x\sin x}.$

Common Mistake

Multiplying the derivatives

It is incorrect to write $dydx=2xcos⁡x\frac{dy}{dx}=2x\cos x$ . The product rule contains two terms joined by addition.

Tip

A useful memory pattern

Write the product rule as $u v^{'} + v u^{'}$ . This compact form helps you remember that each factor is differentiated once.

The quotient rule

A quotient is the result of dividing one expression by another. For example, $x2+1x−1\frac{x^2+1}{x-1}$ is a quotient.

Definition

The quotient rule

y=u(x)v(x),y=\frac{u(x)}{v(x)},

then

dydx=vdudx−udvdxv2,\frac{dy}{dx} = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2},

provided $v≠0v\neq 0$ .

The order of the numerator matters: it is “bottom times derivative of top, minus top times derivative of bottom”.

Example

Differentiating a quotient

Differentiate

y=x2+1x−1.y=\frac{x^2+1}{x-1}.

Let $u=x^2+1$ and $v = x - 1$ . Then
$dudx=2x,dvdx=1.\frac{du}{dx}=2x, \qquad \frac{dv}{dx}=1.$
Apply the quotient rule:
$dydx=(x−1)(2x)−(x2+1)(1)(x−1)2.\frac{dy}{dx} = \frac{(x-1)(2x)-(x^2+1)(1)}{(x-1)^2}.$
Simplify the numerator:
$dydx=2x2−2x−x2−1(x−1)2=x2−2x−1(x−1)2.\begin{aligned} \frac{dy}{dx} &= \frac{2x^2-2x-x^2-1}{(x-1)^2}\\ &= \boxed{\frac{x^2-2x-1}{(x-1)^2}}. \end{aligned}$

Common Mistake

Reversing the subtraction

The numerator is $v u^{'} - u v^{'}$ , not $u v^{'} - v u^{'}$ . Reversing these terms changes the sign of the derivative.

Sometimes a quotient can be rewritten using negative powers. For example, $3x2=3x−2\frac{3}{x^2}=3x^{-2}$ is easier to differentiate using the power rule.

The chain rule

A composite function is a function placed inside another function. In $3x^2-5)^4$ , the inner function is $3x^2-5$ and the outer function raises its input to the fourth power.

Definition

The chain rule

If $y = f (u)$ and $u = g (x)$ , then

dydx=dydududx.\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}.

Equivalently,

ddxf(g(x))=f′(g(x))g′(x).\frac{d}{dx}f(g(x)) = f'(g(x))g'(x).

Differentiate the outer function while keeping the inner expression intact, then multiply by the derivative of the inner function.

Example

Differentiating a composite power

Differentiate $y=(3x^2-5)^4$ .

Set $u=3x^2-5$ , giving $y=u^4$ . Differentiate the outer function:
$dydu=4u3.\frac{dy}{du}=4u^3.$
Differentiate the inner function:
$dudx=6x.\frac{du}{dx}=6x.$
Multiply the two derivatives and replace $u$ :
$dydx=4(3x2−5)3(6x)=24x(3x2−5)3.\frac{dy}{dx} = 4(3x^2-5)^3(6x) = \boxed{24x(3x^2-5)^3}.$

Common Mistake

Forgetting the inner derivative

Writing only $4(3x^2-5)^3$ misses the factor of $6 x$ . Always differentiate the inner function as well.

The chain rule also applies to trigonometric and exponential composites:

ddxsin⁡(g(x))=cos⁡(g(x))g′(x),\frac{d}{dx}\sin(g(x))=\cos(g(x))g'(x),

ddxeg(x)=eg(x)g′(x).\frac{d}{dx}e^{g(x)}=e^{g(x)}g'(x).

Combining the rules

A function may have more than one layer of structure. Work from the main, outermost structure and use another rule when differentiating each part.

Example

Combining product and chain rules

Differentiate $y=x^2(3x+1)^5$ .

The main structure is a product, so choose $u=x^2$ and $v=(3x+1)^5$ . Then $u^{'} = 2 x$ .
Differentiate $v$ using the chain rule:
$v′=5(3x+1)4⋅3=15(3x+1)4.v'=5(3x+1)^4\cdot 3=15(3x+1)^4.$
Apply the product rule:
$dydx=15x2(3x+1)4+2x(3x+1)5.\frac{dy}{dx} = 15x^2(3x+1)^4+2x(3x+1)^5.$
Factorise the common terms:
$dydx=x(3x+1)4(15x+2(3x+1))=x(3x+1)4(21x+2).\begin{aligned} \frac{dy}{dx} &= x(3x+1)^4\left(15x+2(3x+1)\right)\\ &= \boxed{x(3x+1)^4(21x+2)}. \end{aligned}$

Tip

Keep brackets while differentiating

Do not expand complicated powers unless expansion clearly makes the calculation easier. The chain rule often preserves a much simpler form.

Derivatives of inverse functions

An inverse function, written $f^{-1}$ , reverses the action of $f$ . If $y = f (x)$ , then $x=f^{-1}(y)$ .

The notation $f^{-1}(x)$ does not mean $1f(x)\frac{1}{f(x)}$ .

Key Idea

Reciprocal derivative relationship

If $y = f (x)$ and $dydx≠0\frac{dy}{dx}\neq 0$ , then

dxdy=1dydx.\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}.

Therefore,

(f−1)′(y)=1f′(x),y=f(x).\left(f^{-1}\right)'(y)=\frac{1}{f'(x)}, \qquad y=f(x).

The chain rule explains this because $f^{-1}(f(x))=x$ . Differentiating gives

(f−1)′(f(x))f′(x)=1.\left(f^{-1}\right)'(f(x))f'(x)=1.

Example

Finding an inverse derivative at a point

Let $f(x)=x^3+x$ . Find $(f−1)′(2)\left(f^{-1}\right)'(2)$ .

Find the input mapped to 2:
$f(1)=1^3+1=2,$
so $f^{-1}(2)=1$ .
Differentiate the original function:
$f′(x)=3x2+1,f′(1)=4.f'(x)=3x^2+1, \qquad f'(1)=4.$
Take the reciprocal at the corresponding input:
$(f−1)′(2)=14.\boxed{\left(f^{-1}\right)'(2)=\frac{1}{4}}.$

Common Mistake

When the reciprocal rule fails

The formula cannot be used where $f^{'} (x) = 0$ , because division by zero is undefined. An inverse must also exist on the domain being considered.

Connected rates of change

Connected rates of change problems involve two or more quantities that change with time and are linked by an equation.

If $y$ depends on $x$ , and $x$ depends on time $t$ , the chain rule gives

dydt=dydxdxdt.\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}.

A dependency diagram showing the chain rule for connected rates and the reciprocal derivative relationship for inverse functions

Example

Finding the rate of change of a circle's area

The radius of a circle increases at $s−13\text{ cm s}^{-1}$ . Find the rate at which its area increases when the radius is $cm5\text{ cm}$ .

Link area $A$ and radius $r$ :
$A=πr2,dAdr=2πr.A=\pi r^2, \qquad \frac{dA}{dr}=2\pi r.$
Use the given rate $drdt=3\frac{dr}{dt}=3$ in the chain rule:
$dAdt=dAdrdrdt=2πr⋅3.\frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt} = 2\pi r\cdot 3.$
Substitute $r = 5$ :
$s−1.\boxed{\frac{dA}{dt}=30\pi\text{ cm}^2\text{ s}^{-1}}.$

Common Mistake

Substituting too early

Differentiate the relationship before substituting the particular values. If you replace a changing variable with a number too soon, its derivative disappears.

Exam technique

In the exam

Identify the main structure before choosing a differentiation rule.
Write the rule symbolically before substituting the functions or rates.
Keep brackets around composite expressions and simplify only after differentiating.
In connected-rates questions, include the correct units and use signs to show whether a quantity is increasing or decreasing.
Check that a quotient denominator is non-zero and that an inverse derivative does not require division by zero.

Self review

Check yourself

Can you differentiate $xsin⁡xx2+1\frac{x\sin x}{x^2+1}$ using the necessary rules?
What factor is missing from the incorrect derivative $5(2x^3+1)^4$ ?
How would you connect $dVdt\frac{dV}{dt}$ , $dVdr\frac{dV}{dr}$ and $drdt\frac{dr}{dt}$ for a changing sphere?

What you'll learn

How to recognise and use the product, quotient and chain rules.
How to combine several differentiation rules in one calculation.
How derivatives of inverse functions are related.
How the chain rule connects different rates of change.

Before you begin

You should already be able to differentiate powers, exponentials and trigonometric functions. For example,

ddx(xn)=nxn−1,ddx(ex)=ex,ddx(sin⁡x)=cos⁡x.\frac{d}{dx}(x^n)=nx^{n-1}, \qquad \frac{d}{dx}(e^x)=e^x, \qquad \frac{d}{dx}(\sin x)=\cos x.

The new challenge is deciding what to do when functions are multiplied, divided or placed inside one another.

Key Idea

Identify the structure first

Before differentiating, ask whether the function is a product, a quotient or a composite function. Its structure determines the rule you need.

The product rule

A product is the result of multiplying two expressions. For example, $x^2\sin x$ is a product of $x^2$ and $sin⁡x\sin x$ .

You cannot usually differentiate the two factors separately and then multiply their derivatives.

Definition

The product rule

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

dydx=udvdx+vdudx.\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.

In words: differentiate one factor at a time, leaving the other factor unchanged, and add the results.

Example

Differentiating a product

Differentiate $y=x^2\sin x$ .

Choose $u=x^2$ and $v=sin⁡xv=\sin x$ , so
$dudx=2x,dvdx=cos⁡x.\frac{du}{dx}=2x, \qquad \frac{dv}{dx}=\cos x.$
Substitute into the product rule:
$dydx=x2cos⁡x+(sin⁡x)(2x).\frac{dy}{dx} = x^2\cos x+(\sin x)(2x).$
Simplify:
$dydx=x2cos⁡x+2xsin⁡x.\boxed{\frac{dy}{dx}=x^2\cos x+2x\sin x}.$

Common Mistake

Multiplying the derivatives

It is incorrect to write $dydx=2xcos⁡x\frac{dy}{dx}=2x\cos x$ . The product rule contains two terms joined by addition.

Tip

A useful memory pattern

Write the product rule as $u v^{'} + v u^{'}$ . This compact form helps you remember that each factor is differentiated once.

The quotient rule

A quotient is the result of dividing one expression by another. For example, $x2+1x−1\frac{x^2+1}{x-1}$ is a quotient.

Definition

The quotient rule

y=u(x)v(x),y=\frac{u(x)}{v(x)},

then

dydx=vdudx−udvdxv2,\frac{dy}{dx} = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2},

provided $v≠0v\neq 0$ .

The order of the numerator matters: it is “bottom times derivative of top, minus top times derivative of bottom”.

Example

Differentiating a quotient

Differentiate

y=x2+1x−1.y=\frac{x^2+1}{x-1}.

Let $u=x^2+1$ and $v = x - 1$ . Then
$dudx=2x,dvdx=1.\frac{du}{dx}=2x, \qquad \frac{dv}{dx}=1.$
Apply the quotient rule:
$dydx=(x−1)(2x)−(x2+1)(1)(x−1)2.\frac{dy}{dx} = \frac{(x-1)(2x)-(x^2+1)(1)}{(x-1)^2}.$
Simplify the numerator:
$dydx=2x2−2x−x2−1(x−1)2=x2−2x−1(x−1)2.\begin{aligned} \frac{dy}{dx} &= \frac{2x^2-2x-x^2-1}{(x-1)^2}\\ &= \boxed{\frac{x^2-2x-1}{(x-1)^2}}. \end{aligned}$

Common Mistake

Reversing the subtraction

The numerator is $v u^{'} - u v^{'}$ , not $u v^{'} - v u^{'}$ . Reversing these terms changes the sign of the derivative.

Sometimes a quotient can be rewritten using negative powers. For example, $3x2=3x−2\frac{3}{x^2}=3x^{-2}$ is easier to differentiate using the power rule.

The chain rule

A composite function is a function placed inside another function. In $3x^2-5)^4$ , the inner function is $3x^2-5$ and the outer function raises its input to the fourth power.

Definition

The chain rule

If $y = f (u)$ and $u = g (x)$ , then

dydx=dydududx.\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}.

Equivalently,

ddxf(g(x))=f′(g(x))g′(x).\frac{d}{dx}f(g(x)) = f'(g(x))g'(x).

Differentiate the outer function while keeping the inner expression intact, then multiply by the derivative of the inner function.

Example

Differentiating a composite power

Differentiate $y=(3x^2-5)^4$ .

Set $u=3x^2-5$ , giving $y=u^4$ . Differentiate the outer function:
$dydu=4u3.\frac{dy}{du}=4u^3.$
Differentiate the inner function:
$dudx=6x.\frac{du}{dx}=6x.$
Multiply the two derivatives and replace $u$ :
$dydx=4(3x2−5)3(6x)=24x(3x2−5)3.\frac{dy}{dx} = 4(3x^2-5)^3(6x) = \boxed{24x(3x^2-5)^3}.$

Common Mistake

Forgetting the inner derivative

Writing only $4(3x^2-5)^3$ misses the factor of $6 x$ . Always differentiate the inner function as well.

The chain rule also applies to trigonometric and exponential composites:

ddxsin⁡(g(x))=cos⁡(g(x))g′(x),\frac{d}{dx}\sin(g(x))=\cos(g(x))g'(x),

ddxeg(x)=eg(x)g′(x).\frac{d}{dx}e^{g(x)}=e^{g(x)}g'(x).

Combining the rules

A function may have more than one layer of structure. Work from the main, outermost structure and use another rule when differentiating each part.

Example

Combining product and chain rules

Differentiate $y=x^2(3x+1)^5$ .

The main structure is a product, so choose $u=x^2$ and $v=(3x+1)^5$ . Then $u^{'} = 2 x$ .
Differentiate $v$ using the chain rule:
$v′=5(3x+1)4⋅3=15(3x+1)4.v'=5(3x+1)^4\cdot 3=15(3x+1)^4.$
Apply the product rule:
$dydx=15x2(3x+1)4+2x(3x+1)5.\frac{dy}{dx} = 15x^2(3x+1)^4+2x(3x+1)^5.$
Factorise the common terms:
$dydx=x(3x+1)4(15x+2(3x+1))=x(3x+1)4(21x+2).\begin{aligned} \frac{dy}{dx} &= x(3x+1)^4\left(15x+2(3x+1)\right)\\ &= \boxed{x(3x+1)^4(21x+2)}. \end{aligned}$

Tip

Keep brackets while differentiating

Do not expand complicated powers unless expansion clearly makes the calculation easier. The chain rule often preserves a much simpler form.

Derivatives of inverse functions

An inverse function, written $f^{-1}$ , reverses the action of $f$ . If $y = f (x)$ , then $x=f^{-1}(y)$ .

The notation $f^{-1}(x)$ does not mean $1f(x)\frac{1}{f(x)}$ .

Key Idea

Reciprocal derivative relationship

If $y = f (x)$ and $dydx≠0\frac{dy}{dx}\neq 0$ , then

dxdy=1dydx.\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}.

Therefore,

(f−1)′(y)=1f′(x),y=f(x).\left(f^{-1}\right)'(y)=\frac{1}{f'(x)}, \qquad y=f(x).

The chain rule explains this because $f^{-1}(f(x))=x$ . Differentiating gives

(f−1)′(f(x))f′(x)=1.\left(f^{-1}\right)'(f(x))f'(x)=1.

Example

Finding an inverse derivative at a point

Let $f(x)=x^3+x$ . Find $(f−1)′(2)\left(f^{-1}\right)'(2)$ .

Find the input mapped to 2:
$f(1)=1^3+1=2,$
so $f^{-1}(2)=1$ .
Differentiate the original function:
$f′(x)=3x2+1,f′(1)=4.f'(x)=3x^2+1, \qquad f'(1)=4.$
Take the reciprocal at the corresponding input:
$(f−1)′(2)=14.\boxed{\left(f^{-1}\right)'(2)=\frac{1}{4}}.$

Common Mistake

When the reciprocal rule fails

The formula cannot be used where $f^{'} (x) = 0$ , because division by zero is undefined. An inverse must also exist on the domain being considered.

Connected rates of change

Connected rates of change problems involve two or more quantities that change with time and are linked by an equation.

If $y$ depends on $x$ , and $x$ depends on time $t$ , the chain rule gives

dydt=dydxdxdt.\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}.

A dependency diagram showing the chain rule for connected rates and the reciprocal derivative relationship for inverse functions

Example

Finding the rate of change of a circle's area

The radius of a circle increases at $s−13\text{ cm s}^{-1}$ . Find the rate at which its area increases when the radius is $cm5\text{ cm}$ .

Link area $A$ and radius $r$ :
$A=πr2,dAdr=2πr.A=\pi r^2, \qquad \frac{dA}{dr}=2\pi r.$
Use the given rate $drdt=3\frac{dr}{dt}=3$ in the chain rule:
$dAdt=dAdrdrdt=2πr⋅3.\frac{dA}{dt} = \frac{dA}{dr}\frac{dr}{dt} = 2\pi r\cdot 3.$
Substitute $r = 5$ :
$s−1.\boxed{\frac{dA}{dt}=30\pi\text{ cm}^2\text{ s}^{-1}}.$

Common Mistake

Substituting too early

Differentiate the relationship before substituting the particular values. If you replace a changing variable with a number too soon, its derivative disappears.

Exam technique

In the exam

Identify the main structure before choosing a differentiation rule.
Write the rule symbolically before substituting the functions or rates.
Keep brackets around composite expressions and simplify only after differentiating.
In connected-rates questions, include the correct units and use signs to show whether a quantity is increasing or decreasing.
Check that a quotient denominator is non-zero and that an inverse derivative does not require division by zero.

Self review

Check yourself

Can you differentiate $xsin⁡xx2+1\frac{x\sin x}{x^2+1}$ using the necessary rules?
What factor is missing from the incorrect derivative $5(2x^3+1)^4$ ?
How would you connect $dVdt\frac{dV}{dt}$ , $dVdr\frac{dV}{dr}$ and $drdt\frac{dr}{dt}$ for a changing sphere?

What you'll learn

Before you begin

The product rule

The quotient rule

The chain rule

Combining the rules

Derivatives of inverse functions

Connected rates of change

Review 1.10.4 Product, quotient and chain rules (A-level only) by teaching Genie

1.10.4 Product, quotient and chain rules (A-level only)

What you'll learn

Before you begin

The product rule

The quotient rule

The chain rule

Combining the rules

Derivatives of inverse functions

Connected rates of change

Review 1.10.4 Product, quotient and chain rules (A-level only) by teaching Genie

1.10.4 Product, quotient and chain rules (A-level only)

What you'll learn

Before you begin

The product rule

The quotient rule

The chain rule

Combining the rules

Derivatives of inverse functions

Connected rates of change

Review 1.10.4 Product, quotient and chain rules (A-level only) by teaching Genie

1.10.4 Product, quotient and chain rules (A-level only) Revision Guide

1.10.4 Product, quotient and chain rules (A-level only)

What you'll learn

Before you begin

The product rule

The quotient rule

The chain rule

Combining the rules

Derivatives of inverse functions

Connected rates of change

Review 1.10.4 Product, quotient and chain rules (A-level only) by teaching Genie

1.10.4 Product, quotient and chain rules (A-level only) Revision Guide