Differentiation (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

Exam code: 0580 & 0980

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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=x2+3x+5

      • Substitute x=2 in to get y=22+3×2+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 dydx=...

      • pronounced "dy by dx"

      • This refers to the change in y over change in x

    • For example, the gradient function dydx=2x+3

      • Substitute x=2 in to get dydx=2×2+3=7

      • The gradient at the point where x=2 is 7

  • The gradient function dydx 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, dydx=...

  • To differentiate a power of x, bring down the power and reduce the power by 1

    • Differentiating y=x5 gives dydx=5x4

    • Differentiating y=x8 gives dydx=8x7

    • Differentiating y=xn gives dydx=nxn−1

  • Any number that is already in front of x is multiplied by the power that was brought down

    • Differentiating y=2x5 gives dydx=2×5x4=10x4

    • Differentiating y=12x8 gives dydx=12×8x7=4x7

    • Differentiating y=kxn gives dydx=knxn−1

  • Be careful with two special cases:

    • Differentiating y=kx gives dydx=k

      • The x disappears leaving just a number

      • For example, y=2x differentiates to dydx=2

      • This makes sense, the gradient of the straight line y=2x is 2

    • Differentiating just a number (a constant term), y=c, gives dydx=0

      • Numbers disappear!

      • For example, y=4 differentiates to dydx=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 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=x5+x8 gives dydx=5x4+8x7

      • Differentiating y=4x3−10x6 gives dydx=12x2−60x5

  • Remember the two special cases of:

    • kx differentiating to just k

    • and constant terms, c, differentiating to zero

      • E.g. differentiating y=8x+1 gives dydx=8

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

  • 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 dydx equal to that gradient and solve the equation

Examiner Tips and Tricks

Don't forget to write the left-hand sides of y=... and dydx=... 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=x3−2x2+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 dydx=3x2−4x

      • Substitute in x=2 to get dydx=3×22−4×2=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=x3−3x2−4x+1.

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

Find the gradient function using differentiation

dydx=3x2−3×2x−4+0dydx=3x2−6x−4

Substitute x=1 into the gradient function

dydx=3×12−6×1−4=−7

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 dydx=5
(This is not the same as substituting in x=5)

Form an equation using dydx from above

3x2−6x−4=5

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

3x2−6x−9=0x2−2x−3=0(x−3)(x+1)=0x=3 or x=−1

These are the x-coordinates of the two points on the curve with gradient 5
Substitute x=3 into y=x3−3x2−4x+1 to find the y-coordinate of this point

y=33−3×32−4×3+1=−11

Similarly, substitute x=−1 into the equation for y

y=(−1)3−3×(−1)2−4×(−1)+1=1

Write out the two sets of coordinates

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

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What is a gradient function?

A gradient function gives you the gradient of a curve at a point: substitute the point's x-coordinate into it and out comes the gradient there.

  • The equation of the curve, y = ..., gives the y-coordinate when you substitute in x
  • The gradient function, written dy/dx = ... (said "dy by dx"), gives the gradient when you substitute in x
  • dy/dx means the change in y over the change in x
  • It is also called the derivative, or derived function
  • You find it by differentiating the equation of the curve

Example from this page:

  • Curve y = x² + 3x + 5: substituting x = 2 gives y = 2² + 3 × 2 + 5 = 15, so (2, 15) lies on the curve
  • Gradient function dy/dx = 2x + 3: substituting x = 2 gives dy/dx = 2 × 2 + 3 = 7, so the gradient at x = 2 is 7

You need a function rather than a single number because the gradient of a curve changes as you move along it.

How do you find the gradient of a curve at a given point?

Differentiate the equation of the curve to get dy/dx, then substitute the x-coordinate of the point into dy/dx.

  1. Find the x-coordinate of the point (you do not need the y-coordinate)
  2. Differentiate y = ... to get the gradient function dy/dx = ... — multiply by the power, then take one off the power; kx becomes k, and a constant becomes 0
  3. Substitute the x-coordinate into dy/dx

Worked example from this page: find the gradient of y = x³ − 3x² − 4x + 1 at the point (1, −5).

  • dy/dx = 3x² − 6x − 4
  • Substitute x = 1: dy/dx = 3 × 1² − 6 × 1 − 4 = −7
  • The gradient at (1, −5) is −7

You could instead draw a tangent at the point and measure its gradient, but that is only as accurate as your drawing — differentiation is exact. Write "y =" and "dy/dx =" on the left each time so you don't mix up the curve with its gradient function.

How do you find the coordinates of a point on a curve with a given gradient?

Set dy/dx equal to the gradient you are given, solve for x, then substitute each x back into the equation of the curve to find y.

  1. Differentiate to find dy/dx
  2. Set dy/dx = the given gradient — this is not the same as substituting x = that number
  3. Solve the equation for the x-coordinate(s); it is often a quadratic, so bring all terms to one side
  4. Substitute each x into the original curve equation y = ... to find the y-coordinate(s)

Worked example from this page: find the points on y = x³ − 3x² − 4x + 1 with gradient 5.

  • dy/dx = 3x² − 6x − 4
  • Set it equal to 5: 3x² − 6x − 4 = 5
  • Rearrange: 3x² − 6x − 9 = 0, so x² − 2x − 3 = 0
  • Factorise: (x − 3)(x + 1) = 0, so x = 3 or x = −1
  • x = 3 gives y = −11; x = −1 gives y = 1

The points are (3, −11) and (−1, 1).

Mark Curtis

Author: Mark Curtis

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Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

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Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.