Week 5 – SS3 Second Term Mathematics Notes

 WEEK FIVE

• 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

 2.    Let y = 3x²+ 1/x = 3x² + x^-1
    dy/dx = 6x – x^-2 = 6x – 1
x²
3. 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

 2.    y = (1 – 2x³)^1/2 => (1 – 2x³) ^1/2
    Let u =1 – 2x³, hence y = u^1/2
    dy/dx = = ½ u^-1/2 x( –6x²)
            = -3x² u ^{– ½}= -3x²
u^1/2
         -3x²    = -3x²    
         √ u     √(1 – 2x³)

 3.    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
2.    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

 2.    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 – x² 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)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 – t³
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    
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

 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