SUBJECT: MATHEMATICS

CLASS: SS 3

DATE:

TERM: 2nd TERM

REFERENCE TEXT

New General Mathematics for SS book 3 by J.B Channon

Essential Mathematics for SS book 3

Mathematics Exam Focus

Waec and Jamb past Questions

WEEK FIVE

TOPIC: Differentiation of algebraic functions

Differentiation of algebraic functions
Basic rules of differentiation such as sum and difference, product rule, quotient rule
Maximal and minimum application.

Derivative of algebraic functions

Let f, u, v be functions such that

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

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

f(x + x) – f(x) = {u(x+ x) + v(x+ x) – v(x + x) – u(x) – v(x)}

= u (x +x ) – u(x) + v(x +x ) – v(x)

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

Lim f(x + x) – f(x) = U1(x) + V1(x)

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

Examples:Find the derivative of the following

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

Solution

Let y = 2x3 – 5x2 + 2

dy/dx = 6x2 – 10x

Let y = 3x2+ 1/x = 3x2 + x-1

dy/dx = 6x – x-2 = 6x – 1

Let y = 2x3 + 2x2 + 1

dy/dx=6x2 + 4x

Evaluation: 1. If y = 3x4 – 2x3 – 7x + 5. Find dy/dx

2.Findd (8x3 – 5x2 + 6)

Function of a function (chain rule)

Suppose that we know that y is a function of u and that u is a function of x, how do we find the derivative of y with respect to x?

Given that y = f(x) and u = h(x), what is dy/dx?

dy/dx = , this is called the chain rule

Examples

Find the derivative of the following.(a)y = (3x2 – 2)3 (b) y = (1 – 2x3) (c) 5/(6-x2)3

Solution

y = (3x2 – 2)3

Let u = 3x2 – 2

y = (3x2 – 2)3 => y = u3

y = u3

dy/du = 3u2

du/dx = 6x

dy/dx = = 3u2 x 6x

= 18xu2 = 18x(3x2 – 2)2

y = (1 – 2x3)1/2 => (1 – 2x3) 1/2

Let u =1 – 2x3, hence y = u1/2

dy/dx = = ½ u-1/2 x( –6x2)

= -3x2 u – ½= -3x2

u1/2

-3x2 = -3x2

√ u √(1 – 2x3)

y = 5 = 5(6 – x2)-3

(6 – x2)3

Let u = 6 – x2

y = 5(u)–3

dy/du = -15u –4

du/dx = -2x

dy/dx = dy/du X du/dx = -15u-4 x (-2x) = 30x u-4 = 30x (6 – x2)-4

= 30x_

(6 – x2)4

Evaluation:

Given that y = 1 find dy/dx

(2x + 3)4

If y = (3x2 + 1)3 , Find dy/dx

Product Rule

We shall consider the derivative of y = uv where u and v are function of x.

Let y = uv

Then y + y = (u +u )(v + v)

= uv + uv + vu +uv

y = uv + uv + vu+ uv – uv

y= uv + vu + uv

x x

As x =>0 ,u=> 0 , v=> 0

Lim y = Lim uv + Lim vu + Lim uv

x=>0 x x=>0 x x=>0 x x=>0 x

Hence dy/dx= U dv + Vdu

dxdx

Examples

Find the derivatives of the following.

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

Solution

y = (3 + 2x) (1 – x)

Let u = 3 + 2x and v = (1 – x)

du/dx = 2 and dv/dx = -1

dv/dx = u dv + vdu

dx dx

= (1 – x) 2 + (3 + 2x) (-1) = 2 – 2x – 3 – 2x

dy/dx = - 1 – 4x

y = (1 – 2x + 3x2) (4 – 5x2)

Let u = (1 – 2x + 3x2) and v = (4 – 5x2)

du/dx = -2 + 6x and dv/dx = - 10x

dy/dx = udv + vdu

dxdx

= (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

Given that (i) y = (5+3x)(2-x) (ii) y = (1+x)(x+2)3/2 ,find dy/dx

Quotient Rule:

If y = u

then; dy = vdu- udv

dxdxdx

Examples:

Differentiate the following with respect to x. (a)x2 + 1 (b)(x – 1)2

1 – x2 √x

Solution:

y = x2 + 1

1 – x2

Let u = x2 + 1 du/dx = 2x

v= 1 – x2 dv/dx = - 2x

dy = vdu- udv

dxdxdx

dy/dx = (1 – x2)(2x) – (x2 + 1)(-2x)

(1 – x2)2

= 2x – 2x3 + 2x3 + 2x

(1 – x2)2

dy/dx = 4x

– x2)2

y = (x – 1)2

√x

Let u = (x – 1)2 du/dx = 2(x – 1)

v = √x dv/dx = 1/2√x

dy/dx = √x 2(x - 1) -(x – 1)2 1/2√x

(√x)2

dy/dx = √x 2(x - 1) - (x – 1)2 1/2√x

Evaluation: Differentiate with respect to x: (1) (2x + 3)3 (2) √x

(x3– 4)2 √(x + 1)

Applications of differentiation:

There are many applications of differential calculus.

Examples:

Find the gradient of the curve y = x3 – 5x2 + 6x – 3 at the point where x = 3.

Solution:

Y = x3 – 5x2 + 6x – 3

dy/dx = 3x2 – 10x + 6

where x = 3; dy/dx = 3(32) – 10(3) + 6

= 27 – 30 + 6

= 3.

Find the coordinates of the point on the graph of y = 5x2 + 8x – 1 at which the gradient is – 2

Solution:

Y = 5x2 + 8x – 1

dy/dx = 10x + 8

replace; dy/dx by – 2

10x + 8 = - 2

10x = - 2 – 8

x = -10/10 = - 1

Find the point at which the tangent to the curve y = x2 - 4x + 1 at the point (2, -3)

Solution:

Y =x2 - 4x + 1

dy/dx = 2x – 4

at point (2, -3): dy/dx = 2(2) – 4

dy/dx = 0

tangent to the curve: y – y1 = dy/dx(x – x1)

y – (-3) = 0 ( x- 2)

y + 3 = 0

Evaluation:

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

Velocity and Acceleration

Velocity: The velocity after t seconds is the rate of change of displacement with respect to time.

Suppose; s = distance and t = time,

Then; Velocity = ds/dt

Acceleration: This is the rate of change of velocity compared with 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

ds/dt = 8t – 3

velocity = ds/dt = 8(4) – 3

= 32 – 3

= 29

Acceleration: dv/dt = 8.

Maxima and Minimal

Find the maximum and minimum value of y on the curve 6x – x2.

Solution:

y = 6x – x2

dy/dx = 6 – 2x

equatedy/dx = 0

6 – 2x = 0

6 = 2x

X = 3

The turning point is (3, 9)

Find the maximum and minimum of the function x3 – 12x + 2.

Solution:

Y =x3 – 12x + 2

dy/dx = 3x2 – 12

3x2 – 12 = 0

3x2 = 12

x2 = 12/3

x2 = 4 x = ± 2

minimum point occur when d2y/dx2> 0

maximum point occurs when d2y/dx2< 0

d2y/dx2= 6x

substitute x = 2; d2y/dx2 = 6 x 2 = 12

therefore: the function is minimum at point x = 2 and y = - 14

substitute x = - 2; d2y/dx2 = 6(-2) = -12

therefore: the function is maximum at point x = - 2 and y = 18

Evaluation:

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

General Evaluation

Use product rule to find the derivative of

y = x2 (1 + x)½
y = √x (x2 + 3x – 2)2
Find the derivative of y =(7x2 -5)3
Using completing the square method find t if s=ut+1at2

If 3 is a root of the equation x2 – kx +42=0 find the value of k and the other root of the equation

READING ASSIGNMENT: NGM for SS 3 Chapter 10 page 90 -101,

WEEKEND ASSIGNMENT

OBJECTIVE

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 = 3x5wrt x. (a) 15x3 (b) 45 x4 (c) 60x3 (d) 3x5 (e) 12x3

3.If f(x) = 3x2 + 2/x find f1(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

THEORY

Differentiate with respect to x. y2 + x2 – 3xy = 4
Find the derivative of 3x3(x2 + 4)2

Lesson Notes By Weeks and Term - Senior Secondary School 3