Differentiation of Algebraic Function SS3 Mathematics Lesson Note

1. Differentiation of algebraic functions: meaning of differentiation
2. Differentiation from the first principle
3. 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

fl(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 – the 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 concerning x.

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

Example 1

Find from the 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 the 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: 

1. Find from the first principle, the derivatives of y concerning x:

• Y = 3x3                 

1. Y = 7x2 3. Y = 3x2 – 5x

Rules of Differentiation: Let = xn y + dy = (x + dx)n

= xn + nxn-1dx + n(n -1) xn-2(dx)2 + … (dx)n

2!

                         = xn + n xn-1dx + n(n-1) xn-2  (dx)2+ — + (dx)n – xn         2!

                         = nxn-1dx + n (n – 1) xn–1 (dx)2 2!

dy/dx = n xn-1 + n (n –1) xn-1 dx Lim dy/dx = nxn-1 dx = 0

Hence;   dy/dx  =  nxn-1  if y = xn

Example 3:

Find the derivative of the following with respect to x:   (a) x7 (b) x½ (c) 5x2 – 3x (d) – 3x2 (e) y = 2x3 – 3x + 8

Solution

1. Let  y = x7

dy/dx = 7 x7-1 = 7x6

1. Let  y = x ½

dy/dx = ½ x½ -1 = ½ x– ½  =   1

                                                                  2√x

1. Let y = 5x2 – 3x

dy/dx = 10x – 3

1. Let y = – 3x2

dy/dx =2× – 3x2-1 = – 6x

1. Let y = 2x3 – 3x + 8

dy/dx= 3 x 2x3-1 – 3 + 0

= 6x2 – 3

Evaluation:

• If  y=5x4 ,find  dy/dx   
• 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 the idea of the difference between two squares; 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 pages 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 the first principle, the derivative of   y = x + 1/x
2. Find the derivative of 2x2 – 2/x3