Differentiation

What you'll learn

How to differentiate powers of $x$ and terms such as $1x\frac{1}{x}$ .
How to find stationary points by setting $dydx=0\frac{dy}{dx}=0$ .
How to decide where a curve has a negative gradient.
How differentiation connects displacement, velocity, and acceleration.

1. The idea of gradient

The gradient of a straight line tells you how steep it is. For a line, you may remember:

x\text{gradient}=\frac{\text{change in }y}{\text{change in }x}

For a curve, the gradient is not constant. It changes from point to point. To find the gradient at one exact point on a curve, we imagine drawing a tangent.

Definition

Tangent

A tangent is a straight line that touches a curve at one point and has the same direction as the curve at that point.

On a curve, a tangent can slope upwards, be horizontal, or slope downwards. This tells you whether the curve has a positive, zero, or negative gradient at that point.

Diagram showing positive gradient, zero gradient at a stationary point, and negative gradient on a curve

Definition

Derivative

The derivative of $y$ with respect to $x$ , written $dydx\frac{dy}{dx}$ , is the gradient function of a curve.
To differentiate means to find this derivative.

2. The power rule

At IGCSE, the main differentiation rule you need is the power rule.

Key Idea

Power rule

If $y=ax^n$ , then $dydx=anxn−1\frac{dy}{dx}=anx^{n-1}$ . Multiply by the power, then reduce the power by 1.

For example:

$x^3$ differentiates to $3x^2$ .
$5x^4$ differentiates to $20x^3$ .
$7 x$ differentiates to 7.
A constant, such as 11, differentiates to zero.

Example

Differentiating a polynomial

Differentiate $y=5x^4-3x^2+7x-11$ .

Identify each term separately:
$y=5x^4-3x^2+7x-11$
Differentiate $5x^4$ by multiplying by 4 and reducing the power:
$5x4→20x35x^4 \to 20x^3$
Differentiate the remaining terms:
$−3x2→−6x,7x→7,−11→0-3x^2 \to -6x,\qquad 7x \to 7,\qquad -11 \to 0$
Put the differentiated terms together:
$dydx=20x3−6x+7\frac{dy}{dx}=20x^3-6x+7$

Common Mistake

Forgetting the constant

A number on its own disappears when you differentiate. For example, the derivative of $- 11$ is zero, not $- 11$ .

3. Differentiating reciprocal terms

Sometimes you will see a term like $1x\frac{1}{x}$ or $2x\frac{2}{x}$ . Rewrite it using a negative power first:

1x=x−1\frac{1}{x}=x^{-1}

Then use the power rule as normal.

For example:

x−1→−x−2x^{-1} \to -x^{-2}

You can write $x^{-2}$ as $−1x2-\frac{1}{x^2}$ if that is clearer.

Example

Differentiating a curve with a reciprocal term

Differentiate $y=4x2+1xy=4x^2+\frac{1}{x}$ .

Rewrite the reciprocal term using a negative power:
$y=4x^2+x^{-1}$
Differentiate $4x^2$ :
$4x2→8x4x^2 \to 8x$
Differentiate $x^{-1}$ :
$x−1→−x−2x^{-1} \to -x^{-2}$
Write the derivative:
$dydx=8x−x−2\frac{dy}{dx}=8x-x^{-2}$
If preferred, rewrite the negative power as a fraction:
$dydx=8x−1x2\frac{dy}{dx}=8x-\frac{1}{x^2}$

Tip

Negative powers

A negative power does not mean the term is negative. It means the term belongs in the denominator. For example, $x−2=1x2x^{-2}=\frac{1}{x^2}$ .

4. Stationary points

A stationary point is a point on a curve where the gradient is zero. The tangent is horizontal there.

Definition

Stationary point

A stationary point occurs when $dydx=0\frac{dy}{dx}=0$ .

Stationary points can be maximum points, minimum points, or sometimes points of inflection. For many IGCSE questions, you are asked only to find the co-ordinates.

Example

Finding the co-ordinates of a stationary point

The curve $y=4x2+1xy=4x^2+\frac{1}{x}$ has one stationary point. Find its co-ordinates.

Rewrite the equation using a negative power:
$y=4x^2+x^{-1}$
Differentiate:
$dydx=8x−x−2\frac{dy}{dx}=8x-x^{-2}$
Set the derivative equal to zero because the point is stationary:
$8x−1x2=08x-\frac{1}{x^2}=0$
Multiply through by $x^2$ to remove the fraction:
$8x^3-1=0$
Solve for $x$ :
$8x3=1x3=18x=12\begin{aligned} 8x^3&=1\\ x^3&=\frac{1}{8}\\ x&=\frac{1}{2} \end{aligned}$
Substitute $x=12x=\frac{1}{2}$ into the original equation:
$y=4(12)2+112y=4\left(\frac{1}{2}\right)^2+\frac{1}{\frac{1}{2}}$
Calculate $y$ :
$y = 1 + 2 = 3$
Give the co-ordinates:
$(12,3)\left(\frac{1}{2},3\right)$

Common Mistake

Using the derivative for y

After finding the $x$ -coordinate of a stationary point, substitute into the original equation for $y$ , not into $dydx\frac{dy}{dx}$ . The derivative gives the gradient, not the height of the curve.

5. Where a curve has a negative gradient

A curve has a negative gradient when it slopes downwards as $x$ increases.

To find where this happens:

Differentiate to find $dydx\frac{dy}{dx}$ .
Make an inequality: $dydx<0\frac{dy}{dx}<0$ .
Solve the inequality.

Example

Finding the range of values with negative gradient

The curve $C$ has equation $y=x^3-3x^2-9x+6$ . Find the range of values of $x$ for which $C$ has a negative gradient.

Differentiate the equation:
$dydx=3x2−6x−9\frac{dy}{dx}=3x^2-6x-9$
A negative gradient means $dydx<0\frac{dy}{dx}<0$ :
$3x^2-6x-9<0$
Factorise the quadratic:
$3(x^2-2x-3)<0$
Continue factorising:
$3 (x - 3) (x + 1) < 0$
Find the critical values by setting each bracket equal to zero:
$x=3,x=−1x=3,\qquad x=-1$
Since the quadratic opens upwards, it is negative between its two roots:
$- 1 < x < 3$

Tip

Solving gradient inequalities

If $dydx\frac{dy}{dx}$ is an upward-opening quadratic, it is negative between its roots. If it is a downward-opening quadratic, it is negative outside its roots.

Common Mistake

Including the endpoints

For a negative gradient, use strict inequalities. At the endpoints, the gradient is zero, so those values are not included.

6. Differentiation in motion questions

In particle motion questions, displacement is usually written as $s$ , and time is usually written as $t$ .

Definition

Displacement, velocity, and acceleration

Displacement is the position of the particle from a fixed point.
Velocity is the rate of change of displacement: $v=dsdtv=\frac{ds}{dt}$ .
Acceleration is the rate of change of velocity: $a=dvdta=\frac{dv}{dt}$ .

So if you are given $s$ in terms of $t$ :

Differentiate once to get velocity.
Differentiate twice to get acceleration.

Example

Finding when a particle is instantaneously at rest

A particle moves along a straight line. Its displacement from a fixed point is $s=t^3-4t^2-3t+6$ , where $t≥0t\geq0$ . Find the time when the particle is instantaneously at rest.

Instantaneously at rest means the velocity is zero:
$v = 0$
Differentiate displacement to find velocity:
$v=dsdt=3t2−8t−3v=\frac{ds}{dt}=3t^2-8t-3$
Set the velocity equal to zero:
$3t^2-8t-3=0$
Factorise the quadratic:
$(3 t + 1) (t - 3) = 0$
Solve for $t$ :
$t=−13,t=3t=-\frac{1}{3},\qquad t=3$
Use the condition $t≥0t\geq0$ to reject the negative time:
$t = 3$

Common Mistake

At rest does not mean s = 0

“At rest” means the velocity is zero. It does not mean the displacement is zero.

Example

Finding when acceleration has a given value

A particle has displacement $s=t^3-5t^2+7t-1$ . Find the time when its acceleration is 2 m/s².

Differentiate displacement to find velocity:
$v=dsdt=3t2−10t+7v=\frac{ds}{dt}=3t^2-10t+7$
Differentiate velocity to find acceleration:
$a=dvdt=6t−10a=\frac{dv}{dt}=6t-10$
Set the acceleration equal to 2:
$6 t - 10 = 2$
Solve for $t$ :
$6t=12t=2\begin{aligned} 6t&=12\\ t&=2 \end{aligned}$

Exam technique

In the exam

Rewrite fractions such as $1x\frac{1}{x}$ as negative powers before differentiating.
For stationary points or “instantaneously at rest”, set the derivative equal to zero.
For negative gradient questions, solve $dydx<0\frac{dy}{dx}<0$ and use strict inequalities.
In motion questions, remember the chain: displacement $→\to$ velocity $→\to$ acceleration.

Self review

Check yourself

Can you differentiate $y=3x^4-5x^2+7x-9$ ?
If $dydx=(x−2)(x+4)\frac{dy}{dx}=(x-2)(x+4)$ , for which values of $x$ is the gradient negative?
In a motion question, which derivative gives velocity, and which gives acceleration?

Differentiation

What you'll learn

How to differentiate powers of $x$ and terms such as $1x\frac{1}{x}$ .
How to find stationary points by setting $dydx=0\frac{dy}{dx}=0$ .
How to decide where a curve has a negative gradient.
How differentiation connects displacement, velocity, and acceleration.

1. The idea of gradient

The gradient of a straight line tells you how steep it is. For a line, you may remember:

x\text{gradient}=\frac{\text{change in }y}{\text{change in }x}

For a curve, the gradient is not constant. It changes from point to point. To find the gradient at one exact point on a curve, we imagine drawing a tangent.

Definition

Tangent

A tangent is a straight line that touches a curve at one point and has the same direction as the curve at that point.

On a curve, a tangent can slope upwards, be horizontal, or slope downwards. This tells you whether the curve has a positive, zero, or negative gradient at that point.

Diagram showing positive gradient, zero gradient at a stationary point, and negative gradient on a curve

Definition

Derivative

The derivative of $y$ with respect to $x$ , written $dydx\frac{dy}{dx}$ , is the gradient function of a curve.
To differentiate means to find this derivative.

2. The power rule

At IGCSE, the main differentiation rule you need is the power rule.

Key Idea

Power rule

If $y=ax^n$ , then $dydx=anxn−1\frac{dy}{dx}=anx^{n-1}$ . Multiply by the power, then reduce the power by 1.

For example:

$x^3$ differentiates to $3x^2$ .
$5x^4$ differentiates to $20x^3$ .
$7 x$ differentiates to 7.
A constant, such as 11, differentiates to zero.

Example

Differentiating a polynomial

Differentiate $y=5x^4-3x^2+7x-11$ .

Identify each term separately:
$y=5x^4-3x^2+7x-11$
Differentiate $5x^4$ by multiplying by 4 and reducing the power:
$5x4→20x35x^4 \to 20x^3$
Differentiate the remaining terms:
$−3x2→−6x,7x→7,−11→0-3x^2 \to -6x,\qquad 7x \to 7,\qquad -11 \to 0$
Put the differentiated terms together:
$dydx=20x3−6x+7\frac{dy}{dx}=20x^3-6x+7$

Common Mistake

Forgetting the constant

A number on its own disappears when you differentiate. For example, the derivative of $- 11$ is zero, not $- 11$ .

3. Differentiating reciprocal terms

Sometimes you will see a term like $1x\frac{1}{x}$ or $2x\frac{2}{x}$ . Rewrite it using a negative power first:

1x=x−1\frac{1}{x}=x^{-1}

Then use the power rule as normal.

For example:

x−1→−x−2x^{-1} \to -x^{-2}

You can write $x^{-2}$ as $−1x2-\frac{1}{x^2}$ if that is clearer.

Example

Differentiating a curve with a reciprocal term

Differentiate $y=4x2+1xy=4x^2+\frac{1}{x}$ .

Rewrite the reciprocal term using a negative power:
$y=4x^2+x^{-1}$
Differentiate $4x^2$ :
$4x2→8x4x^2 \to 8x$
Differentiate $x^{-1}$ :
$x−1→−x−2x^{-1} \to -x^{-2}$
Write the derivative:
$dydx=8x−x−2\frac{dy}{dx}=8x-x^{-2}$
If preferred, rewrite the negative power as a fraction:
$dydx=8x−1x2\frac{dy}{dx}=8x-\frac{1}{x^2}$

Tip

Negative powers

A negative power does not mean the term is negative. It means the term belongs in the denominator. For example, $x−2=1x2x^{-2}=\frac{1}{x^2}$ .

4. Stationary points

A stationary point is a point on a curve where the gradient is zero. The tangent is horizontal there.

Definition

Stationary point

A stationary point occurs when $dydx=0\frac{dy}{dx}=0$ .

Stationary points can be maximum points, minimum points, or sometimes points of inflection. For many IGCSE questions, you are asked only to find the co-ordinates.

Example

Finding the co-ordinates of a stationary point

The curve $y=4x2+1xy=4x^2+\frac{1}{x}$ has one stationary point. Find its co-ordinates.

Rewrite the equation using a negative power:
$y=4x^2+x^{-1}$
Differentiate:
$dydx=8x−x−2\frac{dy}{dx}=8x-x^{-2}$
Set the derivative equal to zero because the point is stationary:
$8x−1x2=08x-\frac{1}{x^2}=0$
Multiply through by $x^2$ to remove the fraction:
$8x^3-1=0$
Solve for $x$ :
$8x3=1x3=18x=12\begin{aligned} 8x^3&=1\\ x^3&=\frac{1}{8}\\ x&=\frac{1}{2} \end{aligned}$
Substitute $x=12x=\frac{1}{2}$ into the original equation:
$y=4(12)2+112y=4\left(\frac{1}{2}\right)^2+\frac{1}{\frac{1}{2}}$
Calculate $y$ :
$y = 1 + 2 = 3$
Give the co-ordinates:
$(12,3)\left(\frac{1}{2},3\right)$

Common Mistake

Using the derivative for y

After finding the $x$ -coordinate of a stationary point, substitute into the original equation for $y$ , not into $dydx\frac{dy}{dx}$ . The derivative gives the gradient, not the height of the curve.

5. Where a curve has a negative gradient

A curve has a negative gradient when it slopes downwards as $x$ increases.

To find where this happens:

Differentiate to find $dydx\frac{dy}{dx}$ .
Make an inequality: $dydx<0\frac{dy}{dx}<0$ .
Solve the inequality.

Example

Finding the range of values with negative gradient

The curve $C$ has equation $y=x^3-3x^2-9x+6$ . Find the range of values of $x$ for which $C$ has a negative gradient.

Differentiate the equation:
$dydx=3x2−6x−9\frac{dy}{dx}=3x^2-6x-9$
A negative gradient means $dydx<0\frac{dy}{dx}<0$ :
$3x^2-6x-9<0$
Factorise the quadratic:
$3(x^2-2x-3)<0$
Continue factorising:
$3 (x - 3) (x + 1) < 0$
Find the critical values by setting each bracket equal to zero:
$x=3,x=−1x=3,\qquad x=-1$
Since the quadratic opens upwards, it is negative between its two roots:
$- 1 < x < 3$

Tip

Solving gradient inequalities

If $dydx\frac{dy}{dx}$ is an upward-opening quadratic, it is negative between its roots. If it is a downward-opening quadratic, it is negative outside its roots.

Common Mistake

Including the endpoints

For a negative gradient, use strict inequalities. At the endpoints, the gradient is zero, so those values are not included.

6. Differentiation in motion questions

In particle motion questions, displacement is usually written as $s$ , and time is usually written as $t$ .

Definition

Displacement, velocity, and acceleration

Displacement is the position of the particle from a fixed point.
Velocity is the rate of change of displacement: $v=dsdtv=\frac{ds}{dt}$ .
Acceleration is the rate of change of velocity: $a=dvdta=\frac{dv}{dt}$ .

So if you are given $s$ in terms of $t$ :

Differentiate once to get velocity.
Differentiate twice to get acceleration.

Example

Finding when a particle is instantaneously at rest

A particle moves along a straight line. Its displacement from a fixed point is $s=t^3-4t^2-3t+6$ , where $t≥0t\geq0$ . Find the time when the particle is instantaneously at rest.

Instantaneously at rest means the velocity is zero:
$v = 0$
Differentiate displacement to find velocity:
$v=dsdt=3t2−8t−3v=\frac{ds}{dt}=3t^2-8t-3$
Set the velocity equal to zero:
$3t^2-8t-3=0$
Factorise the quadratic:
$(3 t + 1) (t - 3) = 0$
Solve for $t$ :
$t=−13,t=3t=-\frac{1}{3},\qquad t=3$
Use the condition $t≥0t\geq0$ to reject the negative time:
$t = 3$

Common Mistake

At rest does not mean s = 0

“At rest” means the velocity is zero. It does not mean the displacement is zero.

Example

Finding when acceleration has a given value

A particle has displacement $s=t^3-5t^2+7t-1$ . Find the time when its acceleration is 2 m/s².

Differentiate displacement to find velocity:
$v=dsdt=3t2−10t+7v=\frac{ds}{dt}=3t^2-10t+7$
Differentiate velocity to find acceleration:
$a=dvdt=6t−10a=\frac{dv}{dt}=6t-10$
Set the acceleration equal to 2:
$6 t - 10 = 2$
Solve for $t$ :
$6t=12t=2\begin{aligned} 6t&=12\\ t&=2 \end{aligned}$

Exam technique

In the exam

Rewrite fractions such as $1x\frac{1}{x}$ as negative powers before differentiating.
For stationary points or “instantaneously at rest”, set the derivative equal to zero.
For negative gradient questions, solve $dydx<0\frac{dy}{dx}<0$ and use strict inequalities.
In motion questions, remember the chain: displacement $→\to$ velocity $→\to$ acceleration.

Self review

Check yourself

Can you differentiate $y=3x^4-5x^2+7x-9$ ?
If $dydx=(x−2)(x+4)\frac{dy}{dx}=(x-2)(x+4)$ , for which values of $x$ is the gradient negative?
In a motion question, which derivative gives velocity, and which gives acceleration?

Differentiation

What you'll learn

1. The idea of gradient

2. The power rule

3. Differentiating reciprocal terms

4. Stationary points

5. Where a curve has a negative gradient

6. Differentiation in motion questions

Recap questions

You've reached the end

Differentiation

What you'll learn

1. The idea of gradient

2. The power rule

3. Differentiating reciprocal terms

4. Stationary points

5. Where a curve has a negative gradient

6. Differentiation in motion questions

Recap questions

You've reached the end

Differentiation

What you'll learn

1. The idea of gradient

2. The power rule

3. Differentiating reciprocal terms

4. Stationary points

5. Where a curve has a negative gradient

6. Differentiation in motion questions

Recap questions

You've reached the end

Differentiation Revision Guide

Differentiation

What you'll learn

1. The idea of gradient

2. The power rule

3. Differentiating reciprocal terms

4. Stationary points

5. Where a curve has a negative gradient

6. Differentiation in motion questions

Recap questions

You've reached the end

Differentiation Revision Guide