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

1. Differentiation of algebraic functions:
2. Basic rules of differentiation such as sum and difference, product rule, quotient rule
3. 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. Find (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 concerning 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.

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

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 functions 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.

1. 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 to x.  (a)  x2  + 1      (b) (x – 1) 2

                                                                                              1 – x2               √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: 

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 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 equate dy/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:

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 the 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 pages 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