The document defines key concepts related to derivatives including: 1) Notations for derivatives such as F'(x), dy/dx, Dx f(x), and Ϋ. 2) Rules for derivatives including the power rule, product rule, and quotient rule. 3) Examples of applying the rules to find the derivatives of various functions such as f(x) = 4x, f(x) = (x3 - 2x)(x5 + 6x2), and f(x) = (x+2)/x3.

3.  GROUP LEADER : MR.ADEEL IFTIKHAR(20397)
 GROUP MEMBER : MR.FAIZAN FARAZ(20274)
MR.ABDUL HASEEB(20272)

4.  DEFINITION OF DERIVATIVE
 DERIVATIVE NOTATIONS
 DERIVATIVE RULES
 DERIVATIVE OF CONSTANT WITH
EXAMPLE
 POWER RULE WITH EXAMPLE
 PRODUCT RULE WITH EXAMPLE
 QOUTIENT RULE WITH EXAMPLE

5. The derivative of a function of a real
variable measures the sensitivity to
change of a quantity (a function
or dependent variable) which is
determined by another quantity
(the independent variable). It is a
fundamental tool of variable.

6. The derivative of the
position of a moving
object with respect to
time is the
object's velocity

7. These are the notations of derivative.
 F’(x) by Lagrange
 dy/dx by Leibniz
 Dx f(x) or Dxy by Euler
 Ϋ by Newton

8. These are the rules of derivatives.
 The derivative of constant f(x) is always Zero.
 The derivative of x is always 1.
 The Power Rule is subdivided into two parts
1. If f(x) = xⁿ if n is positive then
f’(x)= nxn-1
2. If f(x) = xn if n is negative then
f’(x)= nxn-1
3. If f(x) = xn if n is fraction then
f’(x)= nxn-1

9.  Rules for derivation of addition &
subtraction of functions.
1. If f(x) + g(x)
Then their derivative would be like this
f’(x) + g’(x)
2. If f(x) – g(x)
Then their derivative would be like this
f’(x) – g’(x)

10.  Product Rule
let f(x).g(x)
Then their derivative would be like
= f’(x).g(x) + g’(x).f(x)

11.  The last rule is Quotient Rule.
let suppose h(x) = f(x)/g(x)
Then their derivative would like
dh(x)/dx = { f’(x).g(x) – g’(x).f(x) }/g(x)2

12.  let f(x) = -55
As -55 is a constant number so applying
That derivative of constant function is
always zero so
f’(x) = 0

13.  Let f(x) = 4x0
As we know that any number or variable
raised to power zero is 1.
So x0 = 1
Then f(x) = 4.1 = 4
And f(x) = 4

14. So by derivative of constant function is
always zero.
f’(x) = 0

15. ID = 20272

16. There are also represented by 1st derivative
,2nd derivatives it mean by that how many
times you derivate the functions.
The notations
F’(x) as 1st derivative.
F’’(x) as 2nd derivative.
F’’’(x) as 3rd derivative.
F’’’’(x) as 4th derivative.

17. Derivative can also be represented in terms of
limit as
And limit can be defined as x approaches
to c there is a value L.

18. let f(z) = 4z
So using that derivative of z1 = z
So f(z) = 4.z
f(z) = 4z
Taking derivative on both sides
f’(z) = 4

19. Let f(x) = -3x/4+9 As 4+9 = 13
f(x) = -3x/13
As we know derivative of x1 = 1
But firstly taking derivative on both sides
f’(x) = -3.(dx/dx)/13
So f’(x) = -3/13

20. Let f(x) = x10
Taking derivative on both sides
As power is positive so xn = nxn-1
So f’(x) = 10x10-1
f’(x) = 10x9

21. Let f(x) = -10/x4
Shifting x4 upwards then function will
become
f(x) = -10x-4
As power is negative so derivative will be
f(x) = xn
f’(x) = nxn-1

22. Taking derivative on both sides
f’(x) = -10.(-4.x-4-1)
f’(x) = +40.x-5
So the function in the end will look like
f’(x) = 40/x5

23. Let f(x) = √x5 As in powers (xm)n = xm.n
So the function will become
f(x) = x5/2
By power rule f(x) = xa/b => f’(x) = (a/b)xa/b-1
f’(x) = (a/b)x(a-b)/b

24. So by applying the
f’(x) = (5/2)x5/2 – 1
f’(x) = (5/2)x(5-2)/2 As 5/2 = 2.5
f’(x) = (2.5)x3/2
f’(x) = 2.5x3/2

25. ID :- 20397

26. For derivation it must be kept in mind that
for derivation a function must be
continues. Such that
Left hand limit = Right hand limit

27. A function whose domain remains same
after derivation. Example
f(x) = x3 if 1 < x < 10
And its derivative function will the same
limit
f’(x) = 3x2 1 < x < 10

28. Let f(x) = (x3 – 2x)(x5 + 6x2)
Using power rule of derivative which is
f(x) = g(x).h(x)
Then
f’(x) = g’(x).h(x) + h’(x).g(x)

29. By apply Power Rule
f’(x) = d(x3 – 2x)/dx.(x5 + 6x2) +
d(x5 + 6x2)/dx.(x3 – 2x)
f’(x) = (dx3/dx – d2x/dx).(x5 + 6x2) +
(dx5/dx + d6x2/dx.(x3 – 2x )
f’(x)= {3x2(dx/dx) - 2 (dx/dx)}.(x5 + 6x2) +
{5x4(dx/dx) + 12x(dx/dx)}.(x3 – 2x )

30. f’(x)= (3x2 – 2).(x5 + 6x2) +
(5x4 + 12x).(x3 – 2x)
f‘(x) = {3x2 (x5 + 6x2) -2(x5 + 6x2)} +
{5x4 (x3 – 2x) + 12x(x3 – 2x)}
f’(x) = (3x2.x5 + 18x2.x2 - 2x5 - 12x2 ) +
(5x4.x3 – 10x4.x + 12x.x3 – 24x.x)

31. f’(x) = (3x2.x5 + 18x2.x2 - 2x5 - 12x2 ) +
(5x4.x3 – 10x4.x + 12x.x3 – 24x.x)
f’(x) = (3x2+5 + 18x2+2 - 2x5 - 12x2 ) +
( 5x4+3 – 10x4+1 + 12x1+3 – 24x1+1)
f’(x) = 3x7 + 18x4 - 2x5 - 12x2 +
5x7 – 10x5 + 12x4 – 24x2

32. f’(x) = 3x7 + 18x4 - 2x5 - 12x2 +
5x7 – 10x5 + 12x4 – 24x2
f’(x) = 8x7 + 30x4 - 12x5 - 36x2

33. Let f(x) = (x+2)/x3
Using Quotient rule which is
h(x) = f(x)/g(x)
Then their derivative will be
=> h’(x) = { f’(x).g(x) – g’(x).f(x) } / [f(x)]2

34. By applying Quotient Rule
f’(x) = { [d(x+2)/dx].x3 – [d(x3)/dx].(x+2) } /
(x3)2
f’(x)=[dx/dx + d2/dx].x3 -[3x2.dx/dx].(x+2)/x6
f’(x) = { (1 + 0). x3 - 3x2.(x+2) }/x6
f’(x) = { (1) x3 -3x3 + 6x2}/x6

35. Simplifying the answer
=> f’(x) = { x3 -3x3 + 6x2}/x6
 f’(x) = {-2x3 + 6x2}/x6
 f’(x) = x2.{-2x1 + 6}/x6
 f’(x) = {-2x1 + 6}/x6-2
 f’(x) = (-2x1 + 6)/x4
So the derivative is
f’(x) = (-2x1 + 6)/x4

36. THANK YOU