BAM FORM 5 – DIFFERENTIATION

DIFFERENTIATION BY FIRST PRINCIPLE

Consider a curve that  

Let A(x, f(x)) be a point on the curve let B (x + x, f ( x + x )) be another point on the same curve
  

The gradient of AB

The gradient at A is given by

The gradient at A is given by;

  = limit    

            0

 

Examples

Differentiate the following using first principles

y = x²+3

y = x³+2x²+1

Solution

Solution

Exercise

1) y = x³-x²

2) y = 3x²-2x

3) y = 2x²-4x+1

4) y = 12x²-6x+7

5) y = x²+2x-2

Solutions

 1) y = x³-x²

         

TECHNIQUES OF DIFFERENTIATION

A) DERIVATIVES OF POLYNOMIAL

1) Differentiation of constant [ i.e y = c]

 

                                                                                                                                             

  

∴= 0

2) Differentiation of y = ax where a is a constant

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     

The gradient at B can be found by using the following

y = ax ………..(i)

Taking a point further

y + y = a(x + x)

= ax + ax ……..(ii)

Subtracting y from (iii)

y + y –y = ax + ax – ax

  = 

  = a

3) Differentiation of y = x²
                                                                                                                                                                                                                               

At point B

y + δy

y = [x+ x] ²

= x² + 2x x+ x²  …………… (i)

Subtract y from (i)

y²+ 2xx + x² – x²= y + y – y

y = 2xx+ x²…….

Dividing ….ii,  by x …..

=

As x       0, y    0, y / x = dy /dx

Equation becomes

Note

By proceeding with the same trend we shall get

When y = x⁴, = 4x³

When y = x⁵, = 5x⁴

When y = x⁶, = 6x⁵

Generally

If y = xⁿ , then

      = nx ^n-1

Example

Differentiate the following with respect to x.

  

y = 6x⁷+5x⁴

y =-x¹⁰+9x²

Solution

       
    

 Since differentiation of constant = 0

= 3+0

   = 3

y = 6x⁷+5x⁴-x²

    = 42x⁶+20x³-2x

y =-x¹⁰+9x²

=  

   =  -10x⁹+18x

Exercise

Differentiate the following with respect to x.

i.)     y =5x²+2

ii.)    y =8x³-15x²+6x+2

iii.)  y =x⁵-4x³

iv.) y = 6x²-x³+5x⁴

v.)  y = 3x+10x²-4x⁷

Solution

          y = 5x² + 2
  = 2[5] x^[2-1] +2[0]                                                                                                        

= 10x+0

      = 10x

∴  = 10x

2.)   y = 8x³-15x²+6x+2

           
= 24x²– 30x + 6

3.)   y = x⁵-4x³

        = 5x⁴-12x²

4.)   y=6x²-x³+5x⁴

        =12x-3x²+20x³

DERIVATIVE BY USING CHAIN RULE

Is used to find derivative of function

Example:

Differentiate y = [3x+5]⁴

Solution

Let U = 3x+5

 

y =U⁴

 

4u ³ x3

= 12u³

  = 12 [3x+5]³

Note: This process is called chain rule

2) Find the    if y =

Solution

Let u = x+1

 

y =

y = u ^½

y = ½ u^{– ½}

=   x 

= ½ u^{– ½} x 1

= ½ [x+1]^{– ½}