Calculus: differentiation of simple functions — KCSE Mathematics

KCSE Mathematics · 112 practice questions · 4 syllabus objectives

30 easy30 medium52 hard

What You'll Learn

Key learning outcomes for this topic, aligned to the KNEC KCSE syllabus.

Differentiate polynomials using the power rule: d/dx(xⁿ) = nxⁿ⁻¹; differentiate sums and differences

Find the gradient of a curve y = f(x) at a point by substituting the x-coordinate into f'(x)

Find the equation of the tangent and normal to a curve at a given point using the gradient from differentiation

Calculus: differentiation of simple functions

Sample Questions

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1
easySHORT ANSWER7 marks

In the study of mathematical functions, understanding the rates of change is crucial. The following questions explore the differentiation of various polynomial expressions and their implications in determining gradients and slopes under specific conditions. (a) Differentiate: y=5x^4+3x^3−4. (3 marks) (b) dy/dx when y=5x^2+1/x. (2 marks) (c) Gradient of y=3x²+-5x+3 at x=4. (2 marks)

View Marking Scheme
Part (a) — 3 marks
54x^(4−1) (1 mk)
33x^(3−1) (1 mk)
Constant→0; all correct (1 mk)
Part (b) — 2 marks
1/x=x⁻¹; dy/dx=52x^(2−1)−1x⁻² (1 mk)
Correct expression (1 mk)
Part (c) — 2 marks
dy/dx=x+-5 (1 mk)
At x=4: ×4+-5 (1 mk)
2
easySHORT ANSWER3 marks

Identify the tangent line to the curve y = x^3 - 2x at the point (1, -1). (3 marks)

View Marking Scheme
Part (a) — 3 marks
Differentiate: dy/dx = 3x^2 - 2 (1 mk)
At x=1, dy/dx = 3(1)^2 - 2 = 1 (1 mk)
Use point-slope form: y + 1 = 1(x - 1); tangent equation: y = x - 2 (1 mk)
3
easySHORT ANSWER2 marks

Identify the gradient of the curve y = 3x^2 - 5 at the point x = 2. (2 marks)

View Marking Scheme
Part (a) — 2 marks
Differentiate: dy/dx = 6x (1 mk)
At x=2, dy/dx = 6(2) = 12 (1 mk)
4

Identify the normal to the curve y = x^2 + 2x at the point where x = 0. (4 marks)

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