Differentiation of Algebraic Function II – SS3 Mathematics Lesson Note (Second Term)

Differentiation is the process of finding the derivative dy/dx of a function, which gives its rate of change with respect to x. For algebraic functions, we use the sum and difference rule, the chain rule, the product rule and the quotient rule. The derivative is also used to find gradients, tangents, velocity, acceleration, and maximum and minimum values.

Behavioural objectives

By the end of the lesson, students should be able to:

1. differentiate algebraic functions using the sum and difference rule;
2. differentiate a function of a function using the chain rule;
3. differentiate products and quotients of functions using the product rule and the quotient rule;
4. find the gradient of a curve at a point, the point where a curve has a given gradient, and the equation of a tangent;
5. find velocity and acceleration from a displacement equation;
6. find the maximum and minimum values of a function.

How do we differentiate a sum of algebraic functions?

Let f, u and v be functions such that f(x) = u(x) + v(x). Then:

f(x + δx) = u(x + δx) + v(x + δx)

f(x + δx) – f(x) = u(x + δx) + v(x + δx) – u(x) – v(x)

= [u(x + δx) – u(x)] + [v(x + δx) – v(x)]

Dividing by δx and taking the limit as δx → 0 gives f'(x) = u'(x) + v'(x).

If y = u + v, where u and v are functions of x, then dy/dx = du/dx + dv/dx.

Examples

Find the derivative of the following:

1. 2x³ – 5x² + 2
2. 3x² + 1/x
3. 2x³ + 2x² + 1

Solution

1. Let y = 2x³ – 5x² + 2. Then dy/dx = 6x² – 10x.
2. Let y = 3x² + 1/x = 3x² + x^-1. Then dy/dx = 6x – x^-2 = 6x – 1/x².
3. Let y = 2x³ + 2x² + 1. Then dy/dx = 6x² + 4x.

Evaluation

1. If y = 3x⁴ – 2x³ – 7x + 5, find dy/dx.
2. Find d/dx of (8x³ – 5x² + 6).

What is the chain rule (function of a function)?

Suppose y is a function of u and u is a function of x. How do we find the derivative of y with respect to x?

Given that y = f(u) and u = h(x), then dy/dx = dy/du × du/dx. This is called the chain rule.

Examples

Find the derivative of the following:

1. y = (3x² – 2)³
2. y = (1 – 2x³)
3. y = 5/(6 – x²)³

Evaluation

1. Given that y = 1/(2x + 3)⁴, find dy/dx.
2. If y = (3x² + 1)³, find dy/dx.

What is the product rule?

We consider the derivative of y = uv, where u and v are functions of x. Then:

y + δy = (u + δu)(v + δv) = uv + uδv + vδu + δuδv

δy = uδv + vδu + δuδv

δy/δx = u(δv/δx) + v(δu/δx) + δu(δv/δx)

As δx → 0, δu → 0 and δv → 0. Taking the limit of each term, we have:

dy/dx = u dv/dx + v du/dx

Examples

Find the derivatives of the following:

1. y = (3 + 2x)(1 – x)
2. y = (1 – 2x + 3x²)(4 – 5x²)

Solution

1. y = (3 + 2x)(1 – x)
   
   Let u = 3 + 2x and v = 1 – x. Then du/dx = 2 and dv/dx = -1.
   
   dy/dx = u dv/dx + v du/dx
   
   = (3 + 2x)(-1) + (1 – x)(2)
   
   = -3 – 2x + 2 – 2x
   
   dy/dx = -1 – 4x
2. y = (1 – 2x + 3x²)(4 – 5x²)
   
   Let u = 1 – 2x + 3x² and v = 4 – 5x². Then du/dx = -2 + 6x and dv/dx = -10x.
   
   dy/dx = u dv/dx + v du/dx
   
   = (1 – 2x + 3x²)(-10x) + (4 – 5x²)(-2 + 6x)
   
   = -10x + 20x² – 30x³ + (-8 + 10x² + 24x – 30x³)
   
   = -10x + 20x² – 30x³ – 8 + 10x² + 24x – 30x³
   
   = 14x + 30x² – 60x³ – 8

Evaluation

Find dy/dx given that:

1. y = (5 + 3x)(2 – x)
2. y = (1 + x)(x + 2)^3/2

What is the quotient rule?

If y = u/v, then:

dy/dx = (v du/dx – u dv/dx) / v²

Examples

Differentiate the following with respect to x:

1. (x² + 1)/(1 – x²)
2. (x – 1)²/√x

Evaluation

Differentiate with respect to x:

1. (2x + 3)³/(x³ – 4)²
2. √x/√(x + 1)

How is differentiation applied to gradients and tangents?

Differential calculus has many applications. The examples below use the derivative to find gradients, points on a curve and the equation of a tangent.

Examples

1. Find the gradient of the curve y = x³ – 5x² + 6x – 3 at the point where x = 3.
   
   Solution: dy/dx = 3x² – 10x + 6
   
   When x = 3, dy/dx = 3(3²) – 10(3) + 6 = 27 – 30 + 6 = 3.
2. Find the coordinates of the point on the graph of y = 5x² + 8x – 1 at which the gradient is -2.
   
   Solution: dy/dx = 10x + 8
   
   Replace dy/dx by -2: 10x + 8 = -2
   
   10x = -2 – 8
   
   x = -10/10 = -1
   
   When x = -1, y = 5(-1)² + 8(-1) – 1 = -4. The point is (-1, -4).
3. Find the equation of the tangent to the curve y = x² – 4x + 1 at the point (2, -3).
   
   Solution: dy/dx = 2x – 4
   
   At the point (2, -3): dy/dx = 2(2) – 4 = 0.
   
   The tangent to the curve is y – y₁ = (dy/dx)(x – x₁)
   
   y – (-3) = 0(x – 2)
   
   y + 3 = 0

Evaluation

1. Find the coordinates of the point on the graph of y = x² + 2x – 10 at which the gradient is 8.
2. Find the point on the curve y = x³ + 3x² – 9x + 3 at which the gradient is 15.

What are velocity and acceleration?

Velocity: The velocity after t seconds is the rate of change of displacement with respect to time. If s = distance and t = time, then:

Velocity = ds/dt

Acceleration: This is the rate of change of velocity with respect to time.

Acceleration = dv/dt

Example

A moving body goes s metres in t seconds, where s = 4t² – 3t + 5. Find its velocity after 4 seconds. Show that the acceleration is constant and find its value.

Solution:

s = 4t² – 3t + 5, so ds/dt = 8t – 3.

Velocity = ds/dt = 8(4) – 3 = 32 – 3 = 29.

Acceleration = dv/dt = 8, which is constant.

How do we find maximum and minimum values?

Set dy/dx = 0 to find the turning points. A minimum point occurs when d²y/dx² > 0. A maximum point occurs when d²y/dx² < 0.

Examples

1. Find the maximum and minimum value of y on the curve 6x – x².
   
   Solution: y = 6x – x², so dy/dx = 6 – 2x.
   
   Equate dy/dx = 0: 6 – 2x = 0, so 6 = 2x and x = 3.
   
   The turning point is (3, 9). Since d²y/dx² = -2, which is less than 0, it is a maximum point.
2. Find the maximum and minimum of the function x³ – 12x + 2.
   
   Solution: y = x³ – 12x + 2, so dy/dx = 3x² – 12.
   
   3x² – 12 = 0
   
   3x² = 12
   
   x² = 12/3 = 4
   
   x = ±2
   
   d²y/dx² = 6x
   
   Substitute x = 2: d²y/dx² = 6 × 2 = 12. Therefore the function is a minimum at x = 2 and y = -14.
   
   Substitute x = -2: d²y/dx² = 6(-2) = -12. Therefore the function is a maximum at x = -2 and y = 18.

Evaluation

1. A particle moves in such a way that after t seconds it has gone s metres, where s = 5t + 15t² – t³.
2. Find the maximum and minimum value of y on the curve 4 – 12x – 3x².

General evaluation

Answer the following. Use the product rule for questions 1 and 2.

1. Find the derivative of y = x²(1 + x)^½.
2. Find the derivative of y = √x (x² + 3x – 2)².
3. Find the derivative of y = (7x² – 5)³.
4. Using the completing the square method, find t if s = ut + ½at².
5. If 3 is a root of the equation x² – kx + 42 = 0, find the value of k and the other root of the equation.

Assignment

Reading assignment: NGM for SS 3, Chapter 10, pages 90 to 101.

Weekend assignment: objective questions

1. Differentiate the function 4x⁴ + x³ – 5. (a) 4x³ + 3x² (b) 16x² + 3x² (c) 16x³ + 3x² (d) 16x⁴ + 3x²
2. Find d²y/dx² of the function y = 3x⁵ with respect to x. (a) 15x³ (b) 45x⁴ (c) 60x³ (d) 3x⁵ (e) 12x³
3. If f(x) = 3x² + 2/x, find f'(x). (a) 6x + 2 (b) 6x + 2/x² (c) 6x – 2/x² (d) 6x – 2
4. Find the derivative of 2x³ – 6x². (a) 6x² – 12x (b) 6x² – 12x (c) 2x² – 6x (d) 8x² – 3x
5. Find the derivative of x³ – 7x² + 15x. (a) x² – 7x + 15 (b) 3x² – 14x + 15 (c) 3x² + 7x + 15 (d) 3x² – 7x + 15

Weekend assignment: theory

1. Differentiate with respect to x: y² + x² – 3xy = 4.