WASSCE SHS
The derivative gives the gradient of a curve at a point. For a power of x, multiply by the power and reduce the power by one.
If y = axⁿ, then dy/dx = anxⁿ⁻¹. A constant differentiates to zero. A sum is differentiated term by term. The derivative is a function. The gradient at a point is the number you get when you substitute the x value of that point into the derivative. At a stationary point the gradient is zero, so you set dy/dx = 0 and solve. This page asks for the derivative and one gradient, which is the skill the longer questions build on.
Write the derivative before you substitute. A candidate who substitutes into y and then tries to differentiate a number has nothing left to differentiate. Units are rare in these questions. The answer is a number for a gradient, and an expression for the derivative. Do not cancel carelessly. The derivative of x³ is 3x², not x².
Term by term. The power rule is a mark, and the substitution is a separate mark. A sign error in one term loses that term and may still leave the other term's marks. At x = 2, show 3(2)² rather than jumping to 12. The jump hides whether you squared before you multiplied.
These are practice questions in the style of the paper, with a model answer and a mark scheme. They are not a promise of what your paper will ask.
4 marks
Given y = x³ − 4x, find dy/dx, and find the gradient of the curve at x = 2. (4 marks)
dy/dx = 3x² − 4. At x = 2, gradient = 3 × 4 − 4 = 12 − 4 = 8.
3 marks
Find the x values where the curve y = x³ − 4x has gradient zero. (3 marks)
3x² − 4 = 0 x² = 4/3 x = square root of 4/3 or x = the negative of that square root.
2 marks
State the derivative of a constant, and why the graph of y = 5 has gradient zero everywhere. (2 marks)
The derivative of a constant is 0. The graph is a horizontal line, so the gradient is zero at every point.
More questions on this topic are inside GLearn, marked as you answer, with the reason written out. The first ones need no account.
Practise Elective Mathematics freeMultiply by the old power and reduce the power by one. The derivative of x³ is 3x². The derivative of −4x is −4. A constant on its own becomes zero.
A point where the gradient is zero. Set the derivative equal to zero and solve. On y = x³ − 4x the derivative is 3x² − 4, so the stationary points are where x² = 4/3. All WASSCE Elective Mathematics past questions