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Calculus: differentiation of simple functions — KCSE Mathematics

KCSE Mathematics · 112 practice questions · 4 syllabus objectives · 4 revision lessons

30 easy30 medium52 hard

Last updated · Aligned to the KNEC KCSE syllabus

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

Revision Notes

Concise lesson notes for Calculus: differentiation of simple functions, written to the KCSE Mathematics marking standard. Read the first lesson free below.

Differentiating Polynomials Using Power Rule

In calculus, the power rule is a fundamental technique used to differentiate polynomial functions. The power rule states that for any polynomial term of the form (x^n), the derivative is given by:
[ \frac{d}{dx}(x^n) = nx^{n-1} ]
This means you multiply the term by the exponent and then subtract one from the exponent.
When differentiating sums or differences of polynomial functions, apply the power rule to each term individually.
Example:
Differentiate (f(x) = 3x^4 + 2x^3 - x + 5).

  • For (3x^4): (d/dx(3x^4) = 3 \cdot 4x^{4-1} = 12x^3)
  • For (2x^3): (d/dx(2x^3) = 2 \cdot 3x^{3-1} = 6x^2)
  • For (-x): (d/dx(-x) = -1)
  • For (5): (d/dx(5) = 0)
    Thus, the derivative (f'(x) = 12x^3 + 6x^2 - 1).

Key points to remember

  • Use the power rule: d/dx(x^n) = nx^(n-1).
  • Differentiate each term in sums/differences separately.
  • Constant terms differentiate to zero.

Worked example

Differentiate f(x) = 2x^5 - 4x^2 + 7.

  • f'(x) = 10x^4 - 8x.

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More lessons in this topic

Lesson 2: Finding the Gradient of a Curve

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

To find the gradient of the curve given by the function y = f(x) at a specific point, we first need to calculate the derivative f'(x). The derivative represents the slope of the tangent to the curve at any point.

Steps to find the gradient:

  1. Differentiate the function f(x) to obtain f'(x).
  2. Substitute the x-coordinate of the point into f'(x) to find the gradient at that point.

Example: Let f(x) = 3x^2 + 2x.

  • First, differentiate: f'(x) = 6x + 2.
  • To find the gradient at x = 1, substitute: f'(1) = 6(1) + 2 = 8.

Thus, the gradient of the curve at x = 1 is 8.

  • Differentiate the function to find f'(x).
  • Substitute the x-coordinate into f'(x).
  • The result gives the gradient at that point.

Find the gradient of y = 2x^3 - 4x at x = 2.

  • Differentiate: f'(x) = 6x^2 - 4.
  • Substitute: f'(2) = 6(2^2) - 4 = 20.
Lesson 3: Finding Tangent and Normal Equations

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

To find the equations of the tangent and normal to a curve at a given point, follow these steps:

  1. Differentiate the function to find the gradient (slope) at the point of tangency.
  2. Evaluate the derivative at the specific x-value to obtain the gradient at that point.
  3. Use the point-slope form of the equation of a line:
    • For the tangent: ( y - y_1 = m(x - x_1) )
    • For the normal (which is perpendicular): ( y - y_1 = -\frac{1}{m}(x - x_1) )

Example: Let ( f(x) = x^2 ) and find the equations of the tangent and normal at ( x = 2 ).

  1. Differentiate: ( f'(x) = 2x )
  2. Evaluate at ( x = 2 ): ( f'(2) = 4 ) (gradient of tangent).
  3. Point on curve: ( f(2) = 4 ) (point is (2, 4)).
    • Tangent equation: ( y - 4 = 4(x - 2) ) → ( y = 4x - 4 )
    • Normal equation: ( y - 4 = -\frac{1}{4}(x - 2) ) → ( y = -\frac{1}{4}x + 4.5 )
  • Differentiate the function to find the gradient.
  • Evaluate the derivative at the given point.
  • Use point-slope form for tangent and normal equations.
  • The normal is perpendicular to the tangent.
  • Ensure to simplify the final equations.

Find the tangent and normal to ( f(x) = 3x^2 + 2 ) at ( x = 1 ).

  1. Differentiate: ( f'(x) = 6x ).
  2. Evaluate: ( f'(1) = 6 ) (tangent gradient).
  3. Point on curve: ( f(1) = 5 ) (point (1, 5)).
    • Tangent: ( y - 5 = 6(x - 1) ) → ( y = 6x - 1 ).
    • Normal: ( y - 5 = -\frac{1}{6}(x - 1) ) → ( y = -\frac{1}{6}x + 5.1667 ).
Lesson 4: Differentiation of Simple Functions

Objective: Calculus: differentiation of simple functions

Differentiation is a fundamental concept in calculus used to determine the rate of change of a function. The derivative of a function gives us the slope of the tangent line at any point on the curve. Here are key rules for differentiating simple functions:

  • Power Rule: If f(x) = x^n, then f'(x) = n*x^(n-1).
  • Constant Rule: If f(x) = c (a constant), then f'(x) = 0.
  • Sum Rule: If f(x) = g(x) + h(x), then f'(x) = g'(x) + h'(x).

For example, consider the function f(x) = 3x^3 + 5x^2 - 4. To differentiate:

  1. Apply the Power Rule:
    • For 3x^3, the derivative is 3 * 3x^(3-1) = 9x^2.
    • For 5x^2, the derivative is 2 * 5x^(2-1) = 10x.
    • For -4, the derivative is 0.
  2. Combine the results: f'(x) = 9x^2 + 10x.

Differentiation is crucial in finding maxima and minima in functions, and it helps in understanding motion in physics.

  • The derivative indicates the rate of change of a function.
  • Use the Power Rule for polynomials.
  • Constant functions have a derivative of zero.
  • The Sum Rule allows differentiation of combined functions.

Differentiate f(x) = 4x^4 - 3x + 7.

  • f'(x) = 4 * 4x^(4-1) - 3 * 1 = 16x^3 - 3.

Sample Questions

Read 3 questions and answers free. Sign up to access all 112 questions with full KNEC-style marking schemes and a personalised study plan.

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)

Answer & 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)

Answer & 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)

Answer & 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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Frequently asked questions

What does the KCSE Mathematics topic "Calculus: differentiation of simple functions" cover?

Calculus: differentiation of simple functions covers 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, and more, all aligned to the official KNEC KCSE Mathematics syllabus.

How many practice questions are available for Calculus: differentiation of simple functions?

HighMarks has 112 Calculus: differentiation of simple functions practice questions for KCSE Mathematics, each with a full marking scheme. The first 3 are free; sign up to access the rest, plus all KCSE mock exams and past papers.

Are these aligned with the KNEC KCSE syllabus?

Yes. Every objective on this page is taken directly from the official KNEC KCSE Mathematics syllabus. Practice questions match the KCSE exam format and are graded against the standard KNEC marking scheme.

How should I revise Calculus: differentiation of simple functions for the KCSE exam?

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