BAM FORM 5 – DIFFERENTIATION

Sub topics

-Differentiation by first principles
-Techniques of differentiation
-First and second derivatives
-Implicit differentiation
-Application of differentiation

DIFFERENTIATION BY FIRST PRINCIPLE

The concept of differentiation
The gradient of a curve at a given point is defined as the gradient of the tangent to the curve at that point and is given by the change of y with respect to x.

       As B      A

The gradient of the chord AB
The gradient of a tangent AT at point A

Or

Line  gradient of chord AB = gradient of tangent AT

Example

Find the gradient of the curve

    y = 2x²+ 5

Solution

    At point Q

  y + y = 2 [x+ x]² +5
=2 [x²+ 2x x+ x²] +5
= 2x² +4x x + 2 x² +5…………. (i)

Subtracting y from equation (i)

y + y –y = 2x² + 4xx + 2 x² +5 – [2x² +5]
y = 4xx+ 2²……….(ii)
Dividing ii…………. By x
  = 4x +2x………….iii…

 As x   0, y   0

And

 

 (iii) Becomes

 

 
Note: The expression   is called derivative of y with respect  [w.r.t] to x

The process of finding derivatives is called DIFFERENTIATION.

Example
Differentiate y = x³ +1 with respect to x

Solution
Y +y = [x +x]³ +1
= x³+ 3x²
x + 3xx²+x³+1

Subtracting y
y + y-y = x³+3x²x+3xx²+x³+1 – x³ – 1
y = 3x²x+3x x²+x³

Dividing by x
 

 As x       0,    y                          

   = 3x²

Examples

Find the gradients of the following curves

1.) 2x²-1

2.) y =x³-1

Solution

1)     2x²-1

y+y = 2[x +x]²-1
y+y =2 [x²+2x x+x²]-1
y+y = 2x² +4x x+x²-1…………(i)

Subtracting y from(i)

y + y – y = 2x² + 4xx + x² – 1 – [ 2x²-1]
y = 2x²+ 4xx+2x²-1-2x²+1 ………………..(ii)

Dividing …..(ii) by x

=4x + 2  ………(iii)
As  

(iii) Becomes
 = 4x

1.      y = x³ – 1

Solution

 Subtracting  y
y + y-y = x³ + 3x ² + 3xx² +x³-1-x³ +1

y =3x²x+ 3x+x³

Dividing by
 =  

= 3x² + 3x + x²

 As ,

  = 3x²

EXERCISE

Find the gradients of the following curves.

      1.)     y = x²

Solution

y + ²

             = x² + 2x  + x²                                                         

Subtracting y

^{y + y-y= x2 + 2x  + x2 -x2} 

           y = 2x + x²

Dividing by

 

      = 2x + 

As

 = 2x

2.)    y= x³

Solution

y + x = (x + )³

           ⁼ x³ + 3x² + 3x² + x³…….(i)

Subtracting  y

 y + y-y=X³ + 3x² x + 3xx² + x³ – x³

              y= 3x²x + 3xx² + x³………….(ii)

Dividing by x

 =  

= 3x² + 3xx + ²

As x 0 ,    0 , = 

 x²

       3.)   y = x

solution

y + x = x + x………….(i)

subtracting y

     y + y-y= x + x – x

y= x ………..(ii)

Dividing by  x

 1

4.)  y =3x²                             

Solution

y + x = 3 [x +x] ²

            =3 [x²+ 2x x +x²]

            =3x²+6x x+3x² -3x²……….(i)

Subtracting  y

      y + y-y= 3x²+6xx+3x²-3x²

                 y= 6xx+3x²…….(ii)

Dividing by x

 

= 

As x   0, y      0, = 

 5.) y = x² +3x

Solution

y +x = [x+x]² + 3[x+x]

               x²+2xx+ x²+ 3x+3x…………i

Subtracting y

 y + y-y = X² + 2xx+ x²+3x +3x -(x ² + 3x)

               y = 2xx + x² + 3x……..ii

Dividing by x

 

= 2x +x +3

As x →0,y → 0, →

 2x +3