IGCSE · Practice questions · Differentiation: gradient functions

Down by one.

Extended. Six original Cambridge-style questions on differentiation: the rule for \(ax^n\), gradients at points, and a rate of change in context.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Expand, differentiate each term, substitute.

Prepare: expand any brackets so \(y\) is a sum of terms \(ax^n\).
Rule: \(ax^n \to anx^{n-1}\); \(ax \to a\); a constant \(\to 0\).
Notation: write \(\dfrac{dy}{dx} = \dots\) for the gradient function.
Gradient at a point: substitute the \(x\)-coordinate into \(\dfrac{dy}{dx}\).
Point with a given gradient: solve \(\dfrac{dy}{dx} = k\), then find \(y\) from the original equation.

Recall

[2 marks]

Find \(\dfrac{dy}{dx}\) when (a) \(y = 6x^4\) (b) \(y = \tfrac{1}{2}x^2 - 3x + 4\).

(a) \(6 \times 4x^3 = 24x^3\). ✓
(b) \(\tfrac{1}{2} \times 2x - 3 + 0 = x - 3\). ✓

Recall

[2 marks]

Differentiate \(y = x^3 + 5x^2 - 7\).

\(\dfrac{dy}{dx} = 3x^2 + 10x\). ✓

Calculation

[2 marks]

Find the gradient of the curve \(y = 2x^3 - 4x\) at the point where \(x = -1\).

\(\dfrac{dy}{dx} = 6x^2 - 4\). ✓
At \(x = -1\): \(6(-1)^2 - 4 = 6 - 4 = 2\). ✓

Calculation

[3 marks]

\(y = (x - 2)(x + 5)\). Find the gradient of the curve where it crosses the \(y\)-axis.

Expand: \(y = x^2 + 3x - 10\). ✓
\(\dfrac{dy}{dx} = 2x + 3\). ✓
The curve crosses the \(y\)-axis at \(x = 0\), so the gradient is \(2(0) + 3 = 3\). ✓

Analysis

[3 marks]

The curve \(y = x^2 + kx\) has gradient 10 at the point where \(x = 3\).

(a) Find \(k\). (b) Find the \(x\)-coordinate of the point where the gradient is 0.

(a) \(\dfrac{dy}{dx} = 2x + k\). At \(x = 3\): \(6 + k = 10\), so \(k = 4\). ✓
(b) \(2x + 4 = 0\), so \(x = -2\). ✓

Analysis

[4 marks]

A ball is thrown upwards. Its height, \(h\) metres, after \(t\) seconds is \(h = 30t - 5t^2\).

(a) Find \(\dfrac{dh}{dt}\), the rate of change of height.

(b) Find the rate of change of height when \(t = 1\) and when \(t = 4\), and explain the sign of each answer.

(a) \(\dfrac{dh}{dt} = 30 - 10t\). ✓
(b) \(t = 1\): \(30 - 10 = 20\) m/s, positive, so the ball is rising. ✓
\(t = 4\): \(30 - 40 = -10\) m/s, negative, so the ball is falling. ✓
(c) \(30 - 10t = 0\) gives \(t = 3\) s; \(h = 90 - 45 = 45\) m (its greatest height). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.