Week 2 – SS2 Second Term Further Mathematics Notes

 WEEK TWO
TOPIC: RULES OF DIFFERENTIATION
Derivative of Sum
Let f, U and V be functions of x such that
f(x) = U(x) + V(x)
f(x + x) = U(x + x) + V(x + x)
Therefore f (x + x) – f (x) = {U(x + x) + V(x + x) – U (x) – V(x)}
= U (x) + x) – U(x) + V(x + x) – V(x)
Therefore f (x + x) – f (x) = U(x + x) – U(x) + V(x + x) – V(x)
            x             x        x
Lim f (x + ∆x) – f (x) = Lim U(x + ∆x) – U(x) + LimV(x + ∆x) – V(x)
        ∆x                 x        x
Therefore f‘ (x) = U‘ (x) + V‘(x)
In other words, if y = U + V, where U and V are functions of x, then:
= +
Hence, the derivation of a sum is the sum of the derivatives.
Examples
Find the derivative of each of the following

1. 2x³ – 5x² + 2
2. 3x² +
3. x³ + 2x² + 1

   x

4. + – 3

   Solution

   1. Let y = 2x³ – 5x² + 2

      = 6x² – 10x

   2. Let y = 3x² +

      = 3x² +
      = 6x –

   3. let y = x³ + 2x² + 1

      x= x² + 2x +
      = 2x + 2 –

   4. y = + 1 – 3

      = x + x -3

= x – x
= 1 – 1

 Functions of a Function
Suppose we know that y is a function of u and the u itself is also a functions of x, how do we find the derivation of y with respect to x?
In other words, given y = f (u) and u = h(x)
What is ? By simple re – arrangement we can write
= x ; u 0, x 0
Lim = Lim x }
= lim x lim

 As x      0, u     0
So we can
as lim

 Thus
= x

         x
This is called the chain rule for differentiation.

 Examples
Find the derivative of each of the following:

1. y = (3x² – 2)³
2. y =
3. y = )
4. y = )³

    

5. ²)³

solution

1. Given         y    = (3x² – 2)³

   Let             u    = 3x²– 2
   then        y    = u³

    

    =    6x
    =        
    =    3u² x 6x
    =    18xu2
    =    18x (3x² – 2)²

1.         y    =    

Let     u    =    1-2x³
then    y    =    u

  = u ^{– ½}

 =
= – 6x²
= x
    =        
= )

1.         y    =    ³

   Let         u    =     6 – x²
   then    y    =    
               =    5u^-3
               =    – 15u^-4
               = – 2x
       =        x    
           = -15u^-4 x -2x
           =30 x u^-4
       =    (6-x²)⁴

2.         y    =    )

   Let         u    =    1 + x²

    then    y    =        = 1
           = u         
            = – u
                   1
               = –
           = 2x

             = x
            -1
   = x 2x
   -x
               =

    EVALUATION
   Find the derivative of the followings:
   (i) y = 8x⁵ + 6x – 7
   (ii) y = ( 4x³ – 3)⁴
   (iii) y = 3x² + 1/x³ + 2/x
   The Derivation of a Product
   We shall now 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 – uv
   = uv u + u v
   = u + v+ u
   = u + v
   Examples
   Find the derivative of each of the following

   1. y = (3 + 2x) (1 – x)
   2. y = (1 – 2x + 3x²) (4 – 5x²)
   3. y = (1 + 2x)²
   4. y = x³ (3 – 2x + 4x²)

      Solution

      1. y = (3 + 2x) (1 – x)

         Let u = 3 + 2x; v = 1 – x
         = u + v
         = (3 + 2x) x – 1 + (1 – x) x 2
         = -(3 + 2x) + 2(1 – x)
         = -3 -2x + 2 – 2x
         = – 1 – 4x

      2.     y = (1 – 2x + 3x²)(4 – 5x²)

         Let u + 1 – 2x + 3×2; v =4 – 5 x^{2
             =     -2 + 6x;} = -10x
             =     u + v
             =    (1 -2x + 3x²⁾ x (-10x)
                 + (4 – 5x²)

      3.     y    =     (1 + 2x)²

         Let u     =     ; v = (1 + 2x)²
         
          = ; = 4(1 + 2x)
             = u + v
         = 4(1 + 2x) + (1 + 2x)² x 1
                         
         4 (1 + 2x) + (1 + 2x)²
         

      4. y = x³ (3 – 2x + 4x²)
         

         Let u = x³; v = (3 – 2x + 4x²)
         ⁼ 3x²; = ( – 2 + 8x) x (3 – 2x + 4x²)
         ⁼ 3x²; =
         = u + v

          =     x³ (4x – 1)

            (3x – 2x + 4x^{2) ½ +} 3x²
x (3 – 2x + 4x^{2) ½}
= x³ (4x -1) + 3x² (3 -=2x + 4x^{2 )}
(3x – 2x + 4x² ) ½

 The Derivative of a Quotient
Let y = , where u and v are functions of x and v 0.
y + y =         
        y = = v – u
Examples
Find the derivative of each of the following:

2. 3 + 2x – x²

   

3. 2 + x

   x² + 2x + 7

4.  Solution

   1. y = 1 + x²

      1 – x²
      Let u = 1 + x²; v = 1 – x²
      = 2x; = – 2x
      = v- u
      = (1 – x²) x (2x) – (1 + x²) x (- 2x)
              (1 – x²)²
      = 2x – 2x³ + 2x + 2x³
           (1 – x²)
      = 4x
      (1 – x²)²
      

   2. y = 3 + 2x – x²

      
      Let u = 3 + 2x – x²; = (1 – x)
      = 2 – 2x; = (1 + x)
      = (1 + x) x 2 (1 – x) – (3 + 2x – x²) x
      
      = 4(1 + x)(1 – x) – (3 + 2x – x²)
      2(1 + x)(1 + x)
      = 4 – 4x² – 3 – 2x + x²
      2(1 + x)
      = 1 – 2x – 3x²
      2(1 + x)

   3. y = 2 + x

      x² + 2x + 7
      Put u = 2 + x; v = x² + 2x + 7
      = 1; = 2x + 2
      = (x² + 2x + 7) x 1 – (2 + x)(2x + 2)
      (x² + 2x + 7)²
      = x² + 2x + 7 – 2(x + 2)(x + 1)
      (x² + 2x + 7)²
      = x² + 2x + 7 – 2x² – 6x – 4
      (x² + 2x + 7)²
      = x² + 2x + 7 – 2x² – 6x – 4
      (x² + 2x + 7)²
      = – x² – 4x + 3
      (x² + 2x + 7)²

 HIGHER DERIVATIVES OF THE SECOND AND THIRD ORDER . DIFFERENTIATION OF IMPLICIT FUNCTIONS
HIGHER DERIVATIVES
Given that y = f(x), is also a function of x.
The derivative of with respect to x is
. is called the second derivative of y with respect to x, and it is usually denoted

    (read. Dee two y dee x squared).

 Since      is also a function of x, successive derivatives can be found.

 The third derivative of y with respect to x
Is and is written for short as d³y
dx³
S I d⁴y is the fourth derivative of y with respect to x.
dx⁴
in general is the nth derivative of y with respect to x.
We recall that if y = f(x), is sometimes denoted f¹ (x).
Similarly , , , are sometimes
Denoted fⁿ(x), f^m(x), f^iv (x) respectively.
Example 22
Find the first second and third derivatives of each of the following:

1. 3×4        (b) 3x⁵2x⁴ + x² 1

(c) Inx        (d) e^x4
(e) sin3x²
Solution:
(a)Let y = 3x⁴
Then = 12x³
d²y = 36x²
dx²
d³y = 72x
dx³

1. Let y = 3x⁵ – 2x⁴ + x² -1

   = 15x⁴ – 8x³ + 2x
   d²y = 60x³ – 24x² + 2
   dx²
   

d³y= 180x² – 48x
dx³

1. Let y = Ink

   =
   d²y =
   dx²
   d³y = 2
   dx³x³
   

2. Lat y = e^x4

   = 4x³ e^x4
   

d²y = 12x²e^x4 + 4^x3 e^x4 4^x3
dx²
= 12x²ex⁴+ 16x⁶ – e^x4
d³y= 24 e^x4 + 12x² (e^x4) +96x⁵ e^x
dx³
e^x4 + 16x⁶ (e^x4)
= 24 e^x4 + 48x⁵ e^x4 + 96x⁵ e^x4 + 64x⁹ e^x4

1. Let y = sin3x²

= cos3x² (3x²)
= 6xcos3x²
d²y = 6cos 3x² – 6x sin 3x² (3x²)
dx²
= 6cos 3x² – 36x² sin3x²
d³y = -6sin3x² (3x²) – 72x
dx³
sin3x² 36x²cos3x² (3x²)
= -(6sin3x²) 6x – 17xsin3x²– (36x² cosx²) 6x
= -36xsin3x² -72xsin3x²–     216x³cos3x²
= -108xsin3x₂– 216x³ cos3x²
= -108x (sin3x² ₊ 2x²cos3x²)
EVALUATION
Find the second and third derivatives of (1) cos 6x ( 2) 4x⁵ -5x
Implicit Differentiation
So far, we have treated relations. Of the form y = f (x). Examples of such relations are y = 3x² – 2x + 1, y = 1 +
In any of these relations, y is said to be expressed explicitly in terms of x. The derivative of y with respect to x can be found from the rules of differentiation which have been discussed in the previous units.
Sometimes, the relationship between y and x may not be expressed explicitly.
For example, consider x²y + xy³ + 3x = 0. Here, the relation between y and x is not expressed explicitly. The relationship between y and x is said to be implicit.
In differentiating x²y+xy³+3x=0, y is treated as if it is a function of x and the rules of differentiation are applied in the appropriate manner. The process of differentiating implicit function is called implicit differentiation.

 Examples
Differentiate each of the following implicitly:

1. X² + y² = 4
2. X²y + y² + 4x = 1
3. 4y²x – 5x²y3+ 4y = 0
4. (x + y)² = 5

   Solution
   

5. X² + y² = 4
   

   Differentiating term by term with respect to x:
   2x + 2y = 0
   2y = -2x
   =
   =

6. X² + y²x + 4x = 1
   

   2xy + x² + 2yx + y² + 4 = 0
   (x² + 2yx) = -y² 2xy – 4
   = -(y² + 2xy + 4)
   X² + 2yx

7. 4y²x -5x²y² + 4y = 0

   8yx + 4y² – 15x²y² – 10xy³ + 4 = 0
   (8xy – 15x²y² + 4) = 10xy³ – 4y²
   = 10xy³– 4y²
   8xy – 15x²y² + 4

8. (x + y)² = 5

   x² + 2xy + y² = 5
   2x + 2x +2y+2y = 0
   (2x + 2y) = -2x-2y
   =
   =
   = = -1

 EVALUATION
Differentiate the followings ;
(i) y= (3x+4) (6x-8)
(ii) y = 6x+7/2x-3

 GENERAL EVALUATION
1) Differentiate y = ( 7x⁴ – 6 )⁵
2) Differentiate y = ( 2x + 5) ( 6x – 8)
3) Find the derivative of y = 3x² – 5/x + 3
4) Find the derivative of y = 8/ ( 9 – x⁵)⁴
5) Find the derivative of y = 2x⁴ -5x³ -+ 6
6) If x³– y² + 6xy = 0 find dy/dx
7) Find d³y/dx³ given that y = 8x⁵ – 3x⁴ + 9x³ -7x² +6x+4

 Reading Assignment
New Further MathsProject 2 page 121 – 126

 WEEKEND ASSIGNMENT
1) If y = 3x⁴ -7x + 5 find dy/dx a) 12x³ b) 12x³ – 7 c) 12x³ + 5 d) 12x³ + 12
2) Find the second derivative of cos 5x a) 5sin5x b) -25cos5x c) 25cos5x d) -25sin5x
3) 2) If x²y + 4xy =1 find dy/dx a) 4+2xy/x² b) 4-2xy/x² c) -4-2xy/x² d) -4+2xy/x
4) Given that y = x² + 3x + 2, find dy/dx at x = 2 a) 6 b) 4 c) 7 d) 5
5) Given that y = ( 2x + 3)⁴ find dy/dx a) 18(2x + 3)³ b) 4(2x + 3)⁴ c) 8(2x + 3)³ d) 2( 2x+3)³

 THEORY
1) Differentiate y = (2x² -3)³/x
2) Differentiate y = (2x+ 3)³ (4x² -1)²