Week 4 – SS3 Second Term Mathematics Notes

 WEEK FOUR

• Differentiation of algebraic functions: meaning of differentiation
• Differentiation from first principle
• Standard derivatives of some basic functions.

   

Consider the curve whose equation is given by y = f(x) Recall that m = y₂ – y₁= f(x+x)-f(x)
x₂– x₁x
As point B moves close to A, dx becomes smaller and tends to zero.
The limiting value is written on Lim f(x +x) – f(x) and is denoted by as x –> 0
                    dx
f^l(x) is called the derivative of f(x) and the gradient function of the curve

 The process of finding the derivative of f(x) is called differentiation. The rotations which are commonly used for the derivative of a function are f¹(x) read as f – prime of x, df/dx read as dee x of f
df/dx read dee – f dee- x, dy/dx read dee – y dee- x

 If y = f(x) , this dy/dx = f¹(x) (it is called the differential coefficient of y with respect to x.

 Differentiation from first principle: The process of finding the derivative of a function from the consideration of the limiting value is called differentiation from first principle.

 Example 1
Find from first principle, the derivative of y = x²
Solution
    y = x²
y + y = (x + x)²
y + y = x² + 2xx + (x)²
y = x² + 2xx+ (x)² – y    
y = x² + 2xx + (x)² – x²
y = 2xx + (x)²
y = (2x + x)x
y = 2x + x
x
Lim x = 0
dy = 2x
dx
            
Example 2:
Find from first principle, the derivative of 1/x
Solution
Let y = 1
x
y + y = 1
x + x

 y = 1 – y
x + x
y = 1 – 1
x + x x
y = x – (x + x)
(x +x)x
y = x – x – x
x² + xx
dy = -x
x²+ x
y = -1
x x² + x
Lim x = 0
dy = -1
dx x²

 Evaluation: Find from first principle, the derivatives of y with respect to x:

1. Y = 3x³ 2. Y = 7x² 3. Y = 3x² – 5x

    
Rules of Differentiation: Let    y = xⁿ
y + dy = (x + dx)ⁿ
    = xⁿ + nx^n-1dx + n(n -1) x^n-2(dx)² + … (dx)ⁿ
             2!

  = xⁿ + n x^n-1dx + n(n-1) x^n-2 (dx)²+ — + (dx)ⁿ – xⁿ
             2!
= nx^n-1dx + n (n – 1) x^n–1 (dx)²
            2!
dy/dx = n ^xn-1 + n (n –1) x^n-1 dx
Lim dy/dx = nx^n-1
dx = 0
    Hence; dy/dx = nx^n-1 if y = xⁿ

 Example 3:    
Find the derivative of the following with respect to x: (a) x⁷ (b) x^½ (c) 5x² – 3x (d) – 3x² (e) y = 2x³ – 3x + 8
Solution
a.    Let y = x⁷
    dy/dx = 7 x^7-1 = 7x⁶

 b.    Let y = x ^½
    dy/dx = ½ x^{½ -1} = ½ x^{– ½} = 1
2√x

 c. Let y = 5x² – 3x
dy/dx = 10x – 3
        
d.    Let y = – 3x²
    dy/dx =2× – 3x^2-1 = – 6x
        
e. Let y = 2x³ – 3x + 8
    dy/dx= 3 x 2x^3-1 – 3 + 0
= 6x² – 3

 Evaluation:

1. If y=5x⁴ ,find dy/dx 2.Given that y= 4x^-1 find y¹
   

 General Evaluation
1. Find, from first principles, the derivative of 4x² – 2 with respect to x.
2. Find the derivative of the following a.3x³ – 7x² – 9x + 4 b. 2x³ c. 3/x
3. Using idea of difference of two square; simplify 243x² – 48y²
4. Expand (2x -5)( 3x-4)
5. If the gradient of y=2x²-5 is -12 find the value of y.

 Reading Assignment: NGM for SS 3 Chapter 10 page 82 -88,
Weekend Assignment
Objective
1.    Find the derivative of 5x³(a) 10x² (b) 15x² (c) 10x (d) 15x³
2.    Find dy/dx, if y = 1/x³(a) –3/x⁴    (b) 3/x⁴ (c) 4/x³    (e) –4/x³
3.    Find f¹(x), if f(x) = x³ (a) 3x     (b) 3x²    (c) ½ x³    (d) 2x³
4.    Find the derivative of 1/x(a) 1/x² (b) –1/x² (c) – x     (d) –x²
5.    If y = – 2/3 x³. Find dy/dx (a) 4/3 x² (b) 2x² (c) – 2x² (d) –2x
Theory
1.    Find from first principle, the derivative of y = x + 1/x
2.    Find the derivative of 2x² – 2/x³