# Lesson Notes By Weeks and Term - Senior Secondary School 3

Differentiation of algebraic functions

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

1. Let y = 2x3 – 5x2 + 2

dy/dx = 6x2 – 10x

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

dy/dx = 6x – x-2 = 6x – 1

x2

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)

Dx

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

1. 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

1. 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)

1. 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:

1. Given that y =       1                find dy/dx

(2x + 3)4

1. 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

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

1. 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

1. 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

v

then; dy =  vdu- udv

dxdxdx

v2

Examples:

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

1 – x2               √x

Solution:

1. y = x2 + 1

1 – x2

Let u = x2 + 1      du/dx = 2x

v= 1 – x        dv/dx = - 2x

dy =  vdu-  udv

dxdxdx

v2

dy/dx = (1 – x2)(2x) – (x2 + 1)(-2x)

(1 – x2)2

2x – 2x3 + 2x3 + 2x

(1 – x2)2

dy/dx        =                        4x

1. – x2)2

1. 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

x

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

(x3– 4)2                 √(x + 1)

Applications of differentiation:

There are many applications of differential calculus.

Examples:

1. 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.

1. 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

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:

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.

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

1. 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)

1. 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:

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

General Evaluation

Use product rule to find the derivative of

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

2

1. 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

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