Differentiation

What you'll learn

What $dydx\frac{dy}{dx}$ and $f^{'} (x)$ mean.
How to differentiate powers of $x$ , including roots and fractions.
How to find gradients, tangents, normals and stationary points.
How differentiation helps solve maximum and minimum problems.

1. Gradient and the derivative

You already know that the gradient of a straight line is “change in $y$ divided by change in $x$ ”. A curve is different: its gradient can change from point to point.

To find the gradient of a curve at one point, we use the tangent there: the straight line that touches the curve and has the same gradient at that point.

A curve with a tangent at one point, showing that the derivative is the gradient of that tangent.

Diagram of a curve with tangent line and gradient triangle

Definition

Derivative

The derivative of $y$ with respect to $x$ , written $dydx\frac{dy}{dx}$ , is the function that gives the gradient of the curve at each value of $x$ . If the function is called $f (x)$ , its derivative can also be written as $f^{'} (x)$ .

To differentiate means “find the derivative”.

2. The power rule

Most AS differentiation questions begin with expressions made from powers of $x$ , such as $x^3$ , $x^2$ or $x^{-1}$ .

A coefficient is the number multiplying a power of $x$ . For example, in $5x^3$ , the coefficient is 5.

Key Idea

The power rule

For any constant coefficient $a$ and power $n$ ,

ddx(axn)=anxn−1\frac{d}{dx}(ax^n)=anx^{n-1}

A constant on its own differentiates to 0.

So you multiply by the old power, then reduce the power by 1.

The power rule shown as a simple before-and-after transformation of a term.

Example

Differentiate a polynomial and find a gradient

For $y=3x^3-4x^2+6x-8$ , find $dydx\frac{dy}{dx}$ and the gradient when $x = 2$ .

Differentiate each term using the power rule.
$dydx=9x2−8x+6\frac{dy}{dx}=9x^2-8x+6$
Substitute $x = 2$ into the derivative.
$9(2)^2-8(2)+6=26$
The gradient of the curve when $x = 2$ is 26.

Common Mistake

Substituting too early

Differentiate first, then put in the given $x$ -value. If you put $x = 2$ into $y$ first, you only get a coordinate, not a gradient.

3. Rewriting roots and fractions

The power rule works beautifully if everything is written as a power of $x$ .

Useful rewrites are:

$Common roots and fractions rewritten as powers so the power rule can be used.$

$x=x1/2\sqrt{x}=x^{1/2}$
$1x=x−1\frac{1}{x}=x^{-1}$
$1x2=x−2\frac{1}{x^2}=x^{-2}$
$12x=12x−1\frac{1}{2x}=\frac{1}{2}x^{-1}$

This is why simplifying algebra before differentiating is often the main skill.

Definition

Second derivative

The second derivative, written $d2ydx2\frac{d^2y}{dx^2}$ , is found by differentiating $dydx\frac{dy}{dx}$ . It tells you how the gradient is changing.

Example

Differentiate roots and fractions

For $y=6x+4x+7y=6\sqrt{x}+\frac{4}{x}+7$ , find $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ .

Rewrite the root and fraction as powers.
$y=6x^{1/2}+4x^{-1}+7$
Differentiate once.
$dydx=3x−1/2−4x−2\frac{dy}{dx}=3x^{-1/2}-4x^{-2}$
Differentiate again.
$d2ydx2=−32x−3/2+8x−3\frac{d^2y}{dx^2}=-\frac{3}{2}x^{-3/2}+8x^{-3}$

Tip

Answer form

Negative powers are usually acceptable unless the question asks for a particular form. If you want, $x^{-2}$ can be rewritten as $1x2\frac{1}{x^2}$ .

4. Stationary points, increasing and decreasing

The sign of $dydx\frac{dy}{dx}$ tells you what the curve is doing.

Positive, negative and zero derivative values matched to increasing, decreasing and stationary parts of a curve.

If $dydx>0\frac{dy}{dx}>0$ , the curve is increasing.
If $dydx<0\frac{dy}{dx}<0$ , the curve is decreasing.
If $dydx=0\frac{dy}{dx}=0$ , the curve has a stationary point.

Diagram showing local maximum, local minimum and increasing/decreasing intervals

Definition

Stationary points

A stationary point is a point on a curve where the gradient is zero.
A local maximum is a high point compared with nearby points.
A local minimum is a low point compared with nearby points.

The second derivative can help classify stationary points:

The second derivative test distinguishes a local minimum from a local maximum by the curve’s concavity.

If $d2ydx2>0\frac{d^2y}{dx^2}>0$ , the point is a local minimum.
If $d2ydx2<0\frac{d^2y}{dx^2}<0$ , the point is a local maximum.

Example

Find and classify stationary points

For $y=x^3-6x^2+9x+2$ , find the stationary points and determine their nature.

The cubic in the worked example, showing the local maximum at (1, 6) and local minimum at (3, 2).

Differentiate.
$dydx=3x2−12x+9=3(x−1)(x−3)\frac{dy}{dx}=3x^2-12x+9=3(x-1)(x-3)$
Set the derivative equal to zero.
$\Rightarrow x=1 \text{ or } x=3$
Find the corresponding $y$ -coordinates.
$y(1)=1−6+9+2=6y(3)=27−54+27+2=2\begin{aligned} y(1)&=1-6+9+2=6\\ y(3)&=27-54+27+2=2 \end{aligned}$
Use the second derivative to classify the points.
$minimum\begin{aligned} \frac{d^2y}{dx^2}&=6x-12\\ x=1 &: 6(1)-12=-6<0 \Rightarrow \text{local maximum}\\ x=3 &: 6(3)-12=6>0 \Rightarrow \text{local minimum} \end{aligned}$
The curve has a local maximum at $(1, 6)$ and a local minimum at $(3, 2)$ .
Use the sign of $3 (x - 1) (x - 3)$ to find where the curve is increasing or decreasing.
$1<x<3\begin{aligned} \text{increasing: }&x<1 \text{ or } x>3\\ \text{decreasing: }&1<x<3 \end{aligned}$

Common Mistake

Stationary does not always mean maximum

Solving $dydx=0\frac{dy}{dx}=0$ only finds possible maximum or minimum points. You still need to classify them using the second derivative or a sign-change test.

5. Tangents and normals

A tangent uses the gradient of the curve at a point. A normal is perpendicular to the tangent.

A normal is perpendicular to the tangent at the same point on the curve.

Definition

Tangent and normal

A tangent is a straight line with the same gradient as the curve at a point.
A normal is a straight line perpendicular to the tangent at that point.

For a line through $(a, b)$ with gradient $m$ , use:

y - b = m (x - a)

Then rearrange into the form requested, such as $y = m x + c$ or $a x + b y + c = 0$ .

If the tangent gradient is $m$ , the normal gradient is $−1m-\frac{1}{m}$ .

Common Mistake

Horizontal tangents

If the tangent gradient is 0, the normal is vertical, so its equation is of the form $x = a$ . The formula $−1m-\frac{1}{m}$ would involve division by zero.

Example

Find the equation of a normal

For $y=(2x+1)(x−3)xy=\frac{(2x+1)(x-3)}{x}$ , find the equation of the normal when $x = - 1$ , giving your answer in the form $a x + b y + c = 0$ .

Expand and divide by $x$ first.
$y=2x2−5x−3x=2x−5−3x−1y=\frac{2x^2-5x-3}{x}=2x-5-3x^{-1}$
Differentiate, then evaluate the tangent gradient at $x = - 1$ .
$dydx=2+3x−2,mtangent=2+3(−1)−2=5\frac{dy}{dx}=2+3x^{-2},\qquad m_{\text{tangent}}=2+3(-1)^{-2}=5$
Find the point on the curve.
$y(-1)=2(-1)-5-3(-1)^{-1}=-4$
Find the normal gradient.
$mnormal=−15m_{\text{normal}}=-\frac{1}{5}$
Use the point-gradient form through $(- 1, - 4)$ .
$y+4=−15(x+1)y+4=-\frac{1}{5}(x+1)$
Rearrange into integer form.
$x + 5 y + 21 = 0$

6. Optimisation

Optimisation means finding the value that makes something as large or as small as possible. In differentiation, this usually means finding a maximum or minimum.

Optimisation uses a stationary point of a model curve to find the minimum or maximum value.

Key Idea

Optimisation recipe

Write the quantity to be optimised as a function of one variable.
Differentiate.
Set the derivative equal to zero and solve.
Use the second derivative, or another argument, to justify maximum or minimum.
Substitute back to find the required value.

Example

Minimising a cost model

A journey has cost $C$ pounds when driven at speed $v$ mph, where

C=3200v+2v+15C=\frac{3200}{v}+2v+15

Find the speed that minimises the cost, and find the minimum cost.

Rewrite using powers of $v$ .
$C=3200v^{-1}+2v+15$
Differentiate.
$dCdv=−3200v−2+2\frac{dC}{dv}=-3200v^{-2}+2$
Solve $dCdv=0\frac{dC}{dv}=0$ . Since $v > 0$ , keep the positive root.
$−3200v2+2=0⇒2v2=3200⇒v2=1600⇒v=40-\frac{3200}{v^2}+2=0 \Rightarrow 2v^2=3200 \Rightarrow v^2=1600 \Rightarrow v=40$
Check using the second derivative.
$v=40,d2Cdv2>0\begin{aligned} \frac{d^2C}{dv^2}&=6400v^{-3}\\ \text{when }v=40,\quad \frac{d^2C}{dv^2}&>0 \end{aligned}$
Calculate the minimum cost.
$C=320040+2(40)+15=175C=\frac{3200}{40}+2(40)+15=175$
The minimum cost is £175, achieved at 40 mph.

Tip

When a model has two variables

Use the given constraint, such as a fixed surface area, to rewrite the quantity in one variable before differentiating. You cannot optimise a formula with two changing variables using ordinary AS differentiation until one variable has been eliminated.

Exam technique

In the exam

Simplify first: expand brackets, divide by $x$ , and rewrite roots or fractions as powers.
If you need a gradient, substitute into $dydx\frac{dy}{dx}$ , not into the original $y$ .
For stationary points, always find both coordinates and justify the nature if asked.
For tangent or normal questions, find the point and the gradient before forming the line equation.

Self review

Check yourself

Can you rewrite $x\sqrt{x}$ and $1x3\frac{1}{x^3}$ as powers before differentiating?
If $dydx=0\frac{dy}{dx}=0$ , what tests can you use to decide whether the point is a maximum or a minimum?
For a normal, how do you find its gradient from the tangent gradient?

Differentiation

What you'll learn

What $dydx\frac{dy}{dx}$ and $f^{'} (x)$ mean.
How to differentiate powers of $x$ , including roots and fractions.
How to find gradients, tangents, normals and stationary points.
How differentiation helps solve maximum and minimum problems.

1. Gradient and the derivative

You already know that the gradient of a straight line is “change in $y$ divided by change in $x$ ”. A curve is different: its gradient can change from point to point.

To find the gradient of a curve at one point, we use the tangent there: the straight line that touches the curve and has the same gradient at that point.

A curve with a tangent at one point, showing that the derivative is the gradient of that tangent.

Diagram of a curve with tangent line and gradient triangle

Definition

Derivative

To differentiate means “find the derivative”.

2. The power rule

Most AS differentiation questions begin with expressions made from powers of $x$ , such as $x^3$ , $x^2$ or $x^{-1}$ .

A coefficient is the number multiplying a power of $x$ . For example, in $5x^3$ , the coefficient is 5.

Key Idea

The power rule

For any constant coefficient $a$ and power $n$ ,

ddx(axn)=anxn−1\frac{d}{dx}(ax^n)=anx^{n-1}

A constant on its own differentiates to 0.

So you multiply by the old power, then reduce the power by 1.

The power rule shown as a simple before-and-after transformation of a term.

Example

Differentiate a polynomial and find a gradient

For $y=3x^3-4x^2+6x-8$ , find $dydx\frac{dy}{dx}$ and the gradient when $x = 2$ .

Differentiate each term using the power rule.
$dydx=9x2−8x+6\frac{dy}{dx}=9x^2-8x+6$
Substitute $x = 2$ into the derivative.
$9(2)^2-8(2)+6=26$
The gradient of the curve when $x = 2$ is 26.

Common Mistake

Substituting too early

Differentiate first, then put in the given $x$ -value. If you put $x = 2$ into $y$ first, you only get a coordinate, not a gradient.

3. Rewriting roots and fractions

The power rule works beautifully if everything is written as a power of $x$ .

Useful rewrites are:

$Common roots and fractions rewritten as powers so the power rule can be used.$

$x=x1/2\sqrt{x}=x^{1/2}$
$1x=x−1\frac{1}{x}=x^{-1}$
$1x2=x−2\frac{1}{x^2}=x^{-2}$
$12x=12x−1\frac{1}{2x}=\frac{1}{2}x^{-1}$

This is why simplifying algebra before differentiating is often the main skill.

Definition

Second derivative

The second derivative, written $d2ydx2\frac{d^2y}{dx^2}$ , is found by differentiating $dydx\frac{dy}{dx}$ . It tells you how the gradient is changing.

Example

Differentiate roots and fractions

For $y=6x+4x+7y=6\sqrt{x}+\frac{4}{x}+7$ , find $dydx\frac{dy}{dx}$ and $d2ydx2\frac{d^2y}{dx^2}$ .

Rewrite the root and fraction as powers.
$y=6x^{1/2}+4x^{-1}+7$
Differentiate once.
$dydx=3x−1/2−4x−2\frac{dy}{dx}=3x^{-1/2}-4x^{-2}$
Differentiate again.
$d2ydx2=−32x−3/2+8x−3\frac{d^2y}{dx^2}=-\frac{3}{2}x^{-3/2}+8x^{-3}$

Tip

Answer form

Negative powers are usually acceptable unless the question asks for a particular form. If you want, $x^{-2}$ can be rewritten as $1x2\frac{1}{x^2}$ .

4. Stationary points, increasing and decreasing

The sign of $dydx\frac{dy}{dx}$ tells you what the curve is doing.

Positive, negative and zero derivative values matched to increasing, decreasing and stationary parts of a curve.

If $dydx>0\frac{dy}{dx}>0$ , the curve is increasing.
If $dydx<0\frac{dy}{dx}<0$ , the curve is decreasing.
If $dydx=0\frac{dy}{dx}=0$ , the curve has a stationary point.

Diagram showing local maximum, local minimum and increasing/decreasing intervals

Definition

Stationary points

A stationary point is a point on a curve where the gradient is zero.
A local maximum is a high point compared with nearby points.
A local minimum is a low point compared with nearby points.

The second derivative can help classify stationary points:

The second derivative test distinguishes a local minimum from a local maximum by the curve’s concavity.

If $d2ydx2>0\frac{d^2y}{dx^2}>0$ , the point is a local minimum.
If $d2ydx2<0\frac{d^2y}{dx^2}<0$ , the point is a local maximum.

Example

Find and classify stationary points

For $y=x^3-6x^2+9x+2$ , find the stationary points and determine their nature.

The cubic in the worked example, showing the local maximum at (1, 6) and local minimum at (3, 2).

Differentiate.
$dydx=3x2−12x+9=3(x−1)(x−3)\frac{dy}{dx}=3x^2-12x+9=3(x-1)(x-3)$
Set the derivative equal to zero.
$\Rightarrow x=1 \text{ or } x=3$
Find the corresponding $y$ -coordinates.
$y(1)=1−6+9+2=6y(3)=27−54+27+2=2\begin{aligned} y(1)&=1-6+9+2=6\\ y(3)&=27-54+27+2=2 \end{aligned}$
Use the second derivative to classify the points.
$minimum\begin{aligned} \frac{d^2y}{dx^2}&=6x-12\\ x=1 &: 6(1)-12=-6<0 \Rightarrow \text{local maximum}\\ x=3 &: 6(3)-12=6>0 \Rightarrow \text{local minimum} \end{aligned}$
The curve has a local maximum at $(1, 6)$ and a local minimum at $(3, 2)$ .
Use the sign of $3 (x - 1) (x - 3)$ to find where the curve is increasing or decreasing.
$1<x<3\begin{aligned} \text{increasing: }&x<1 \text{ or } x>3\\ \text{decreasing: }&1<x<3 \end{aligned}$

Common Mistake

Stationary does not always mean maximum

Solving $dydx=0\frac{dy}{dx}=0$ only finds possible maximum or minimum points. You still need to classify them using the second derivative or a sign-change test.

5. Tangents and normals

A tangent uses the gradient of the curve at a point. A normal is perpendicular to the tangent.

A normal is perpendicular to the tangent at the same point on the curve.

Definition

Tangent and normal

A tangent is a straight line with the same gradient as the curve at a point.
A normal is a straight line perpendicular to the tangent at that point.

For a line through $(a, b)$ with gradient $m$ , use:

y - b = m (x - a)

Then rearrange into the form requested, such as $y = m x + c$ or $a x + b y + c = 0$ .

If the tangent gradient is $m$ , the normal gradient is $−1m-\frac{1}{m}$ .

Common Mistake

Horizontal tangents

If the tangent gradient is 0, the normal is vertical, so its equation is of the form $x = a$ . The formula $−1m-\frac{1}{m}$ would involve division by zero.

Example

Find the equation of a normal

For $y=(2x+1)(x−3)xy=\frac{(2x+1)(x-3)}{x}$ , find the equation of the normal when $x = - 1$ , giving your answer in the form $a x + b y + c = 0$ .

Expand and divide by $x$ first.
$y=2x2−5x−3x=2x−5−3x−1y=\frac{2x^2-5x-3}{x}=2x-5-3x^{-1}$
Differentiate, then evaluate the tangent gradient at $x = - 1$ .
$dydx=2+3x−2,mtangent=2+3(−1)−2=5\frac{dy}{dx}=2+3x^{-2},\qquad m_{\text{tangent}}=2+3(-1)^{-2}=5$
Find the point on the curve.
$y(-1)=2(-1)-5-3(-1)^{-1}=-4$
Find the normal gradient.
$mnormal=−15m_{\text{normal}}=-\frac{1}{5}$
Use the point-gradient form through $(- 1, - 4)$ .
$y+4=−15(x+1)y+4=-\frac{1}{5}(x+1)$
Rearrange into integer form.
$x + 5 y + 21 = 0$

6. Optimisation

Optimisation means finding the value that makes something as large or as small as possible. In differentiation, this usually means finding a maximum or minimum.

Optimisation uses a stationary point of a model curve to find the minimum or maximum value.

Key Idea

Optimisation recipe

Write the quantity to be optimised as a function of one variable.
Differentiate.
Set the derivative equal to zero and solve.
Use the second derivative, or another argument, to justify maximum or minimum.
Substitute back to find the required value.

Example

Minimising a cost model

A journey has cost $C$ pounds when driven at speed $v$ mph, where

C=3200v+2v+15C=\frac{3200}{v}+2v+15

Find the speed that minimises the cost, and find the minimum cost.

Rewrite using powers of $v$ .
$C=3200v^{-1}+2v+15$
Differentiate.
$dCdv=−3200v−2+2\frac{dC}{dv}=-3200v^{-2}+2$
Solve $dCdv=0\frac{dC}{dv}=0$ . Since $v > 0$ , keep the positive root.
$−3200v2+2=0⇒2v2=3200⇒v2=1600⇒v=40-\frac{3200}{v^2}+2=0 \Rightarrow 2v^2=3200 \Rightarrow v^2=1600 \Rightarrow v=40$
Check using the second derivative.
$v=40,d2Cdv2>0\begin{aligned} \frac{d^2C}{dv^2}&=6400v^{-3}\\ \text{when }v=40,\quad \frac{d^2C}{dv^2}&>0 \end{aligned}$
Calculate the minimum cost.
$C=320040+2(40)+15=175C=\frac{3200}{40}+2(40)+15=175$
The minimum cost is £175, achieved at 40 mph.

Tip

When a model has two variables

Exam technique

In the exam

Simplify first: expand brackets, divide by $x$ , and rewrite roots or fractions as powers.
If you need a gradient, substitute into $dydx\frac{dy}{dx}$ , not into the original $y$ .
For stationary points, always find both coordinates and justify the nature if asked.
For tangent or normal questions, find the point and the gradient before forming the line equation.

Self review

Check yourself

Can you rewrite $x\sqrt{x}$ and $1x3\frac{1}{x^3}$ as powers before differentiating?
If $dydx=0\frac{dy}{dx}=0$ , what tests can you use to decide whether the point is a maximum or a minimum?
For a normal, how do you find its gradient from the tangent gradient?

Differentiation

What you'll learn

1. Gradient and the derivative

2. The power rule

3. Rewriting roots and fractions

4. Stationary points, increasing and decreasing

5. Tangents and normals

6. Optimisation

Recap questions

You've reached the end

Differentiation

What you'll learn

1. Gradient and the derivative

2. The power rule

3. Rewriting roots and fractions

4. Stationary points, increasing and decreasing

5. Tangents and normals

6. Optimisation

Recap questions

You've reached the end

Differentiation

What you'll learn

1. Gradient and the derivative

2. The power rule

3. Rewriting roots and fractions

4. Stationary points, increasing and decreasing

5. Tangents and normals

6. Optimisation

Recap questions

You've reached the end

Differentiation Revision Guide

Differentiation

What you'll learn

1. Gradient and the derivative

2. The power rule

3. Rewriting roots and fractions

4. Stationary points, increasing and decreasing

5. Tangents and normals

6. Optimisation

Recap questions

You've reached the end

Differentiation Revision Guide