Differentiation (Edexcel IGCSE Maths A (Modular))

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Written by: Mark Curtis

Reviewed by: Dan Finlay

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Differentiation

What is a gradient function?

Recall that the equation of a curve gives the y-coordinate of a point when you substitute in its x-coordinate
- For example, $y = x^{2} + 3 x + 5$
  - Substitute $x = 2$ in to get $y = 2^{2} + 3 \times 2 + 5 = 15$
  - The point $(2, 15)$ lies on the curve
A gradient function gives the gradient of the curve at a point when you substitute in its x-coordinate
- A gradient function is written as $\frac{d y}{d x} = . . .$
  - pronounced "dy by dx"
  - This refers to the change in y over change in x
- For example, the gradient function $\frac{d y}{d x} = 2 x + 3$
  - Substitute $x = 2$ in to get $\frac{d y}{d x} = 2 \times 2 + 3 = 7$
  - The gradient at the point where $x = 2$ is 7
The gradient function $\frac{d y}{d x}$ can also be called the derivative or derived function

What is differentiation and how does it work?

Differentiation is an algebraic method that changes the equation of a curve, $y = . . .$ , into a gradient function, $\frac{d y}{d x} = . . .$
To differentiate a power of $x$ , bring down the power and reduce the power by 1
- Differentiating $y = x^{5}$ gives $\frac{d y}{d x} = 5 x^{4}$
- Differentiating $y = x^{8}$ gives $\frac{d y}{d x} = 8 x^{7}$
- Differentiating $y = x^{n}$ gives $\frac{d y}{d x} = n x^{n - 1}$
Any number that is already in front of $x$ is multiplied by the power that was brought down
- Differentiating $y = 2 x^{5}$ gives $\frac{d y}{d x} = 2 \times 5 x^{4} = 10 x^{4}$
- Differentiating $y = \frac{1}{2} x^{8}$ gives $\frac{d y}{d x} = \frac{1}{2} \times 8 x^{7} = 4 x^{7}$
- Differentiating $y = k x^{n}$ gives $\frac{d y}{d x} = k n x^{n - 1}$
Be careful with two special cases:
- Differentiating $y = k x$ gives $\frac{d y}{d x} = k$
  - The $x$ disappears leaving just a number
  - For example, $y = 2 x$ differentiates to $\frac{d y}{d x} = 2$
  - This makes sense, the gradient of the straight line $y = 2 x$ is 2
- Differentiating just a number (a constant term), $y = c$ , gives $\frac{d y}{d x} = 0$
  - Numbers disappear!
  - For example, $y = 4$ differentiates to $\frac{d y}{d x} = 0$
  - This makes sense, the gradient of the horizontal line $y = 4$ is zero

Image showing how the equation y = kx^n differentiates to dy/dx = knx^(n-1)

How do I differentiate negative powers of x?

The same rules apply to negative powers, $y = x^{n}$ becomes $\frac{d y}{d x} = n x^{n - 1}$
- Be careful: subtracting 1 from a negative power creates a larger negative number!
  - E.g. $y = x^{- 3}$ becomes $\frac{d y}{d x} = - 3 x^{- 3 - 1} = - 3 x^{- 4}$
You may need to use index laws to convert algebraic fractions into negative powers (and vice versa)
- For example
  - $y = \frac{1}{x} \to y = x^{- 1}$ so $\frac{d y}{d x} = - x^{- 2} = - \frac{1}{x^{2}}$
  - $y = \frac{5}{x^{6}} \to y = 5 x^{- 6}$ so $\frac{d y}{d x} = 5 \times (- 6) x^{- 7} = - \frac{30}{x^{7}}$
  - $y = \frac{4}{3 x^{10}} \to y = \frac{4}{3} x^{- 10}$ so $\frac{d y}{d x} = \frac{4}{3} \times (- 10) x^{- 11} = - \frac{40}{3 x^{11}}$

How do I differentiate sums and differences of terms?

The equation of a curve may include a number of different terms
- You can differentiate each term individually
  - Differentiating $y = x^{5} + x^{8}$ gives $\frac{d y}{d x} = 5 x^{4} + 8 x^{7}$
  - Differentiating $y = 4 x^{3} - 10 x^{6}$ gives $\frac{d y}{d x} = 12 x^{2} - 60 x^{5}$
Remember the two special cases of:
- $k x$ differentiating to just $k$
- and constant terms, $c$ , differentiating to zero
  - E.g. differentiating $y = 8 x + 1$ gives $\frac{d y}{d x} = 8$

Image showing the equation y = 3x^2 - 5x + 3 differentiates to dy/dx = 6x - 5

You may need to separate an algebraic fraction
- E.g. $y = \frac{x^{3} + 2 x}{x^{5}}$ becomes $y = \frac{x^{3}}{x^{5}} + \frac{2 x}{x^{5}} = \frac{1}{x^{2}} + \frac{2}{x^{4}}$
  - Using negative indices gives $y = x^{- 2} + 2 x^{- 4}$
  - This can be differentiated: $\frac{d y}{d x} = - 2 x^{- 3} + 2 (- 4) x^{- 5}$
  - This simplifies to $\frac{d y}{d x} = - \frac{2}{x^{3}} - \frac{8}{x^{5}}$

How do I find the gradient of a curve using the gradient function?

Find the x-coordinate of the point on the curve you're interested in
Use differentiation to turn the equation of the curve, $y = . . .$ , into the gradient function, $\frac{d y}{d x} = . . .$
Substitute the x-coordinate into the gradient function to find the gradient
- The y-coordinate is not needed

Image showing how the gradient of the curve y = 3x^2 - 2x at the point (2, 8), by differentiating the function and then substituting in x = 2. The gradient of the function at this (2, 8) is 10.

If instead you are given a gradient and asked to find the x-coordinate
- set $\frac{d y}{d x}$ equal to that gradient and solve the equation

Examiner Tips and Tricks

Don't forget to write the left-hand sides of $y = . . .$ and $\frac{d y}{d x} = . . .$ to avoid mixing up the curve equation with the gradient function!

The gradient of a curve changes as you move along the curve
- To find the gradient at a particular point you can draw a tangent and find its gradient
- This is a graphical method that is not accurate
  - It depends on how well you draw the tangent
Instead, you can use differentiation to find the gradient function, then substitute the x-coordinate of the point into the gradient function to find the gradient
- This is an algebraic method that is exact
For example, to find the gradient of the curve $y = x^{3} - 2 x^{2} + 6$ at the point P where $x = 2$
- either try to draw a tangent at $x = 2$ and measure its gradient (see below)
- or differentiate the equation to get $\frac{d y}{d x} = 3 x^{2} - 4 x$
  - Substitute in $x = 2$ to get $\frac{d y}{d x} = 3 \times 2^{2} - 4 \times 2 = 4$
  - The gradient is 4

Graph showing the curve with equation y = x^3 - 2x^2 + 6 and a tangent to the curve at a point P.

Worked Example

A curve has the equation $y = x^{3} - 3 x^{2} - 4 x + 1$ .

(a) Find the gradient of the curve at the point $(1, - 5)$ .

Find the gradient function using differentiation

$\begin{array}{rcl} \frac{d y}{d x} & = & 3 x^{2} - 3 \times 2 x - 4 + 0 \\ \frac{d y}{d x} & = & 3 x^{2} - 6 x - 4 \end{array}$

Substitute $x = 1$ into the gradient function

$\begin{array}{rcl} \frac{d y}{d x} & = & 3 \times 1^{2} - 6 \times 1 - 4 \\ = & - 7 \end{array}$

The gradient when $x = 1$ is -7
(The y-coordinate of the point is not needed)

The gradient at $(1, - 5)$ is -7

(b) Find the coordinates of the two points on the curve with a gradient of 5.

This is saying that $\frac{d y}{d x} = 5$
(This is not the same as substituting in $x = 5$ )

Form an equation using $\frac{d y}{d x}$ from above

$3 x^{2} - 6 x - 4 = 5$

This is a quadratic equation
Bring the terms to one side and solve (for example, by factorisation)

$\begin{array}{rcl} 3 x^{2} - 6 x - 9 & = & 0 \\ x^{2} - 2 x - 3 & = & 0 \\ (x - 3) (x + 1) & = & 0 \\ x & = & 3 or x = - 1 \end{array}$

These are the $x$ -coordinates of the two points on the curve with gradient 5
Substitute $x = 3$ into $y = x^{3} - 3 x^{2} - 4 x + 1$ to find the $y$ -coordinate of this point

$\begin{array}{rcl} y & = & 3^{3} - 3 \times 3^{2} - 4 \times 3 + 1 \\ = & - 11 \end{array}$

Similarly, substitute $x = - 1$ into the equation for $y$

$\begin{array}{rcl} y & = & {(- 1)}^{3} - 3 \times {(- 1)}^{2} - 4 \times (- 1) + 1 \\ = & 1 \end{array}$

Write out the two sets of coordinates

The points $(3, - 11)$ and $(- 1, 1)$ have a gradient of 5

Last updated: 18 October 2024

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Differentiation (Edexcel IGCSE Maths A (Modular))

Revision Note

Differentiation

What is a gradient function?

What is differentiation and how does it work?

How do I differentiate negative powers of x?

How do I differentiate sums and differences of terms?

How do I find the gradient of a curve using the gradient function?

How is the gradient function related to drawing tangents?

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Number

Prime Factors, HCF & LCM

Prime Factor Decomposition

Uses of Prime Factor Decomposition

HCF & LCM

Standard Form

Converting To & From Standard Form

Operations with Standard Form

Using Percentages

Percentage Increases & Decreases

Reverse Percentages

Compound Interest & Depreciation

Compound Interest

Depreciation

Ratio Toolkit

Introduction to Ratios

Sharing in a Ratio

Working with Proportion

Ratio Problem Solving

Ratios & FDP

Multiple Ratios

Direct & Inverse Proportion

Direct Proportion

Inverse Proportion

Algebra & Sequences

Rearranging Formulas

Formulas where Subject Appears Once

Formulas where Subject Appears Twice

Algebraic Proof

Algebraic Proof

Solving Inequalities

Solving Linear Inequalities

Solving Quadratic Inequalities

Simultaneous Equations

Linear Simultaneous Equations

Quadratic Simultaneous Equations

Forming & Solving Equations

Forming Equations from Shapes

Problem Solving with Equations

Sequences

Introduction to Sequences

Arithmetic Sequences

Sum of an Arithmetic Series

Functions, Graphs & Differentiation

Functions

Introduction to Functions

Domain & Range

Composite Functions

Inverse Functions

Graphing Inequalities

Representing Inequalities as Regions

Finding Inequalities from Regions

Transformations of Graphs

Translations of Graphs

Reflections of Graphs

Stretches of Graphs

Differentiation & Gradients

Finding Gradients of Tangents

Differentiation

Finding Stationary Points & Turning Points

Classifying Stationary Points

Problem Solving with Differentiation

Using Differentiation for Kinematics

Geometry

Angles in Polygons & Parallel Lines

Basic Angle Properties

Angles in Polygons

Angles in Parallel Lines