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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.12: Algebra and graphs — Differentiation

The notation \(\dfrac{dy}{dx}\) is required. In this syllabus item, the power \(n\) is a positive integer or zero only (and simple sums of at most three such terms). You are not expected to identify points of inflection.

Gradient from a tangent

The gradient of a curve at a point is the gradient of the tangent at that point. Estimate it by drawing the tangent, then using rise/run.

Method

  1. Draw a straight tangent that just touches the curve at the given point.
  2. Pick two convenient points on the tangent and form a right-angled triangle.
  3. Gradient \(= \dfrac{\text{rise}}{\text{run}}\). That value is \(\dfrac{dy}{dx}\) at that \(x\).

Estimate the gradient of \(y = \tfrac{1}{4}x^2\) at \(x = 2\).

Tangent to the parabola at (2, 1) with rise 1 and run 1 giving gradient 1
Rise \(1\), run \(1\), so gradient \(= 1\). The rule \(\dfrac{dy}{dx} = \tfrac{1}{2}x\) gives the same value at \(x = 2\).

The power rule

If \(y = ax^n\), then \(\dfrac{dy}{dx} = nax^{n-1}\). A constant differentiates to \(0\); \(y = 3x\) differentiates to \(3\). Differentiate sums term by term.

Differentiate \(y = x^3 + 2x^2\).

Term by term: derivative of x cubed plus 2 x squared is 3 x squared plus 4 x
\(\dfrac{d}{dx}(x^3) = 3x^2\) and \(\dfrac{d}{dx}(2x^2) = 4x\).

Stationary points

A stationary point is where \(\dfrac{dy}{dx} = 0\). Solve that equation for \(x\), then substitute into \(y\) to get the coordinates.

Method

  1. Find \(\dfrac{dy}{dx}\) and set it equal to \(0\).
  2. Factor or solve the resulting equation. Keep exact fractions.
  3. Substitute each \(x\) into the original \(y\) to get the point \((x, y)\).

Find the stationary points of \(y = x^3 + 2x^2\).

Stationary points at (0, 0) and (negative 4 over 3, 32 over 27)
\(3x^2 + 4x = 0 \Rightarrow x(3x+4)=0\). Then \(y\left(-\dfrac{4}{3}\right) = \dfrac{32}{27}\).

Paper 2 (non-calculator)

Keep fractions: \(x = -\dfrac{4}{3}\), not \(1.33\). For \(y\left(-\dfrac{4}{3}\right)\), write \(\left(-\dfrac{4}{3}\right)^3 + 2\left(-\dfrac{4}{3}\right)^2 = -\dfrac{64}{27} + \dfrac{32}{9} = -\dfrac{64}{27} + \dfrac{96}{27} = \dfrac{32}{27}\).

Maximum or minimum

After you have a stationary point you must classify it. Any one of these is accepted:

  • an accurate sketch of the curve;
  • the second derivative: \(\dfrac{d^2y}{dx^2} > 0\) is a local minimum, \(\dfrac{d^2y}{dx^2} < 0\) is a local maximum;
  • the sign of \(\dfrac{dy}{dx}\) either side of the point (positive then negative is a max; negative then positive is a min).

If the second derivative is \(0\), the test is inconclusive at IGCSE — do not call the point an inflection.

Classify the stationary points of \(y = x^3 + 2x^2\).

Second derivative 6x plus 4 is positive at x equals 0 and negative at x equals negative 4 over 3
\(\dfrac{d^2y}{dx^2} = 6x + 4\). At \(x = 0\): \(4 > 0\) local min. At \(x = -\dfrac{4}{3}\): \(-4 < 0\) local max.
Curve y equals x cubed plus 2 x squared with local maximum at negative 4 over 3 and local minimum at the origin
The sketch agrees: a local maximum at \(\left(-\dfrac{4}{3}, \dfrac{32}{27}\right)\) and a local minimum at \((0, 0)\).

Try this

Find \(\dfrac{dy}{dx}\) for \(y = 4x^3 - 3x + 2\), then the stationary values of \(x\).

Show answer
Answer

  1. Differentiate term by term

    \[ \dfrac{dy}{dx} = 12x^2 - 3 \]

  2. Stationary when the derivative is zero

    \[ 12x^2 - 3 = 0 \Rightarrow 12x^2 = 3 \Rightarrow x^2 = \dfrac{1}{4} \]

  3. Two solutions

    \[ x = \pm \dfrac{1}{2} \]

Exam Traps

  • \(\dfrac{dy}{dx} = 0\) finds a stationary point — you must still classify it as max or min.
  • If the second derivative is \(0\), the test is inconclusive. Do not call the point an inflection (not required at IGCSE).

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