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2e · E2.12–E2.13

Differentiation and functions

Topic 2 Algebra and graphs · Extended only · Papers 2 and 4

EXTENDED Everything on this page is Extended content only. Core candidates are not examined on it.

🎯What you need to be able to do

  • Differentiate \( ax^{n} \) (with \(n\) a positive integer or zero), and sums of up to three such terms, using \( \dfrac{dy}{dx} \) notation.
  • Use the derivative to find the gradient of a curve at a point, and stationary (turning) points.
  • Decide whether a turning point is a maximum or a minimum, by any valid method.
  • Use function notation, and find domains and ranges.
  • Find the inverse function \( f^{-1}(x) \) and composite functions such as \( gf(x) = g(f(x)) \).

📚The mathematics

Differentiation

\( \dfrac{dy}{dx} \) is the gradient function: put in an \(x\)-value and it gives the gradient of the curve (the gradient of the tangent) there. The one rule you need:

\[ y = ax^{n} \;\Longrightarrow\; \frac{dy}{dx} = anx^{n - 1} \]

Multiply by the power, then reduce the power by one. A term in \(x\) becomes a constant (\( 7x \to 7 \)), and a constant disappears. Differentiate a sum term by term.

Stationary points

At a turning point the gradient is zero, so solve \( \dfrac{dy}{dx} = 0 \), then substitute each \(x\) back into the original equation for \(y\). To decide the nature, use any one of:

  • Second derivative: differentiate again. \( \dfrac{d^{2}y}{dx^{2}} < 0 \) ⇒ maximum; \( > 0 \) ⇒ minimum.
  • Gradient either side: \( + \) then \( - \) is a maximum; \( - \) then \( + \) is a minimum.
  • An accurate sketch of the curve.
The cubic y = x cubed minus 6x squared plus 9x plus 2 with a maximum at (1, 6) and a minimum at (3, 2). Small horizontal tangent lines are drawn at both points, and the sign of the gradient is marked: positive before x = 1, negative between 1 and 3, positive after 3.
The worked example: gradient \( + \), \( 0 \), \( - \) around the maximum, and \( - \), \( 0 \), \( + \) around the minimum.

Functions

\( f(x) = 2x - 3 \) means “double, then subtract 3”; \( f(4) \) is the output when the input is 4. The domain is the set of inputs, the range the set of outputs.

  • Composite \( gf(x) = g(f(x)) \): do \(f\) first, then \(g\) — read from the right. Usually \( fg \ne gf \).
  • Inverse \( f^{-1} \) undoes \(f\): write \( y = f(x) \), rearrange for \(x\), then swap \(x\) and \(y\). A useful check: \( f^{-1}(f(a)) = a \).
A mapping diagram for the composite function gf. The input 2 goes through f(x) = 2x minus 3 to 1, then through g(x) = x squared plus 1 to 2. The input 4 goes to 5 and then to 26. The inverse arrow from 5 back to 4 is drawn dashed and labelled f inverse.
\( gf(x) \): apply \(f\) first. The inverse runs the first arrow backwards.

✏️Worked example (non-calculator)

(a) Find the coordinates of the turning points of \( y = x^{3} - 6x^{2} + 9x + 2 \) and determine their nature. [6] (b) \( f(x) = 2x - 3 \) and \( g(x) = x^{2} + 1 \). Find \( fg(2) \), \( gf(x) \) and \( f^{-1}(x) \). [4]

(a) \( \dfrac{dy}{dx} = 3x^{2} - 12x + 9 = 3(x - 1)(x - 3) = 0 \), so \( x = 1 \) or \( x = 3 \). \( y(1) = 1 - 6 + 9 + 2 = 6 \) and \( y(3) = 27 - 54 + 27 + 2 = 2 \). Then \( \dfrac{d^{2}y}{dx^{2}} = 6x - 12 \): at \( x = 1 \) it is \( -6 < 0 \), so \( (1, 6) \) is a maximum; at \( x = 3 \) it is \( 6 > 0 \), so \( (3, 2) \) is a minimum.

(b) \( g(2) = 5 \), so \( fg(2) = f(5) = 7 \). \( gf(x) = (2x - 3)^{2} + 1 \). For the inverse, \( y = 2x - 3 \Rightarrow x = \dfrac{y + 3}{2} \), so \( f^{-1}(x) = \dfrac{x + 3}{2} \).

Check it. \( f^{-1}(f(4)) = f^{-1}(5) = \tfrac{8}{2} = 4 \) ✓. And the gradient just left of \( x = 1 \), at \( x = 0 \), is \( +9 \), while at \( x = 2 \) it is \( -3 \): \( + \) then \( - \), a maximum ✓.
Wrong order. \( fg(2) \) means \(g\) first. Doing \(f\) first gives \( g(f(2)) = g(1) = 2 \), a different answer — which is exactly why the order matters.

📝Practise

All Extended. Questions in the style of the current Paper 2 (non-calculator) and Paper 4 (calculator).

1. (Non-calculator.) Find \( \dfrac{dy}{dx} \) when \( y = 4x^{3} - 5x^{2} + 7x - 2 \). [2]
\( 12x^{2} - 10x + 7 \).
2. (Non-calculator.) Find the gradient of \( y = x^{2} - 5x + 1 \) at the point where \( x = 4 \). [2]
\( \dfrac{dy}{dx} = 2x - 5 = 3 \) at \( x = 4 \).
3. (Non-calculator.) The curve \( y = x^{2} + kx \) has gradient 1 when \( x = 3 \). Find \(k\). [2]
\( 2x + k = 1 \) at \( x = 3 \): \( 6 + k = 1 \), \( k = -5 \).
4. (Non-calculator.) Find the turning points of \( y = x^{3} - 12x \) and determine their nature. [5]
\( 3x^{2} - 12 = 0 \Rightarrow x = \pm 2 \). Points \( (2, -16) \) and \( (-2, 16) \). \( \dfrac{d^{2}y}{dx^{2}} = 6x \): at \( x = 2 \), \( 12 > 0 \), minimum; at \( x = -2 \), \( -12 < 0 \), maximum.
5. (Non-calculator.) Find the equation of the tangent to \( y = 2x^{2} - 3x + 1 \) at the point where \( x = 1 \). [3]
At \( x = 1 \), \( y = 0 \). Gradient \( 4x - 3 = 1 \). Tangent: \( y - 0 = 1(x - 1) \), i.e. \( y = x - 1 \).
6. (Non-calculator.) \( f(x) = 5 - 3x \). Find \( f(-2) \) and \( f^{-1}(x) \). [3]
\( f(-2) = 11 \). \( y = 5 - 3x \Rightarrow x = \dfrac{5 - y}{3} \), so \( f^{-1}(x) = \dfrac{5 - x}{3} \).
7. (Non-calculator.) \( f(x) = \dfrac{3}{x + 2} \) and \( g(x) = 2x - 1 \). Find \( fg(x) \), and write \( gf(x) \) as a single fraction. [4]
\( fg(x) = \dfrac{3}{(2x - 1) + 2} = \dfrac{3}{2x + 1} \). \( gf(x) = \dfrac{6}{x + 2} - 1 = \dfrac{6 - (x + 2)}{x + 2} = \dfrac{4 - x}{x + 2} \).
8. (Non-calculator.) \( h(x) = x^{2} + 3 \) with domain \( \{-2, -1, 0, 1, 2\} \). Write down the range of \(h\). [1]
\( h(\pm 2) = 7 \), \( h(\pm 1) = 4 \), \( h(0) = 3 \): the range is \( \{3, 4, 7\} \). Repeated outputs are listed once.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Desmos — plot a curve and its derivative; the derivative is zero exactly where the curve turns
  • Corbettmaths — composite and inverse functions