# Lesson Notes By Weeks and Term - Senior Secondary School 3

Differentiation of algebraic functions: meaning of differentiation

SUBJECT: MATHEMATICS

CLASS:  SS 3

DATE:

TERM: 2nd TERM

REFERENCE TEXT

• New General Mathematics for SS book 3 by J.B Channon
• Essential Mathematics for SS book 3
• Mathematics Exam Focus
• Waec and Jamb past Questions

WEEK FOUR

TOPIC: Differentiation of algebraic functions: meaning of differentiation

• 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

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 – 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 + 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        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