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INTRODUCTION
GROUP LEADER : MR.ADEEL IFTIKHAR(20397)
GROUP MEMBER : MR.FAIZAN FARAZ(20274) MR.ABDUL HASEEB(20272)
AGENDA FOR TODAY DEFINITION OF DERIVATIVE DERIVATIVE NOTATIONS DERIVATIVE RULES DERIVATIVE OF CONSTANT WITH
EXAMPLE POWER RULE WITH EXAMPLE PRODUCT RULE WITH EXAMPLE QOUTIENT RULE WITH EXAMPLE
DERIVATIVE DEFINITION
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.
EXAMPLE OF DERIVATIVE
The derivative of the position of a moving object with respect to time is the object's velocity
DERIVATIVE NOTATIONS
These are the notations of derivative. F’(x) by Lagrange dy/dx by Leibniz Dx f(x) or Dxy by Euler Ϋ by Newton
RULES OF DERIVATIVE
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 parts1. If f(x) = xⁿ if n is positive then f’(x)= nxn-1
2. If f(x) = xn if n is negative thenf’(x)= nxn-1
3. If f(x) = xn if n is fraction thenf’(x)= nxn-1
CONTINUED
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)
CONTINUED
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
Example of Constant functions
let f(x) = -55As -55 is a constant number so applying That derivative of constant function is always zero so
f’(x) = 0
Example of Constant functions
Let f(x) = 4x0
As we know that any number or variable raised to power zero is 1.So x0 = 1Then f(x) = 4.1 = 4 And f(x) = 4
Number of DerivationsThere 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.
Derivative in terms of limitDerivative can also be represented in terms of limit as
And limit can be defined as x approaches to c there is a value L.
Examples of derivative of x1
let f(z) = 4zSo using that derivative of z1 = zSo f(z) = 4.z f(z) = 4zTaking derivative on both sides f’(z) = 4
Example of derivative of x1
Let f(x) = -3x/4+9 As 4+9 = 13 f(x) = -3x/13As we know derivative of x1 = 1But firstly taking derivative on both sides f’(x) = -3.(dx/dx)/13So f’(x) = -3/13
Examples of Power Rule
Let f(x) = x10
Taking derivative on both sidesAs power is positive so xn = nxn-1
So f’(x) = 10x10-1
f’(x) = 10x9
Examples of Power Rule
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
Continued
Taking derivative on both sidesf’(x) = -10.(-4.x-4-1)f’(x) = +40.x-5
So the function in the end will look like
f’(x) = 40/x5
Example of Power Rule
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
Continued
So by applying thef’(x) = (5/2)x5/2 – 1
f’(x) = (5/2)x(5-2)/2 As 5/2 = 2.5f’(x) = (2.5)x3/2
f’(x) = 2.5x3/2
Continuity & Differentiability
For derivation it must be kept in mind that for derivation a function must be continues. Such that
Left hand limit = Right hand limit
Derivative Functions
A function whose domain remains same after derivation. Examplef(x) = x3 if 1 < x < 10And its derivative function will the same limitf’(x) = 3x2 1 < x < 10
Examples of Product Rule
Let f(x) = (x3 – 2x)(x5 + 6x2)Using power rule of derivative which isf(x) = g(x).h(x)Then f’(x) = g’(x).h(x) + h’(x).g(x)
Continued
By apply Power Rulef’(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 )
Continued
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)
Continued
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
Continued
f’(x) = 3x7 + 18x4 - 2x5 - 12x2 + 5x7 – 10x5 + 12x4 – 24x2
f’(x) = 8x7 + 30x4 - 12x5 - 36x2
Example of Quotient rule
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
Continued
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
Continued
Simplifying the answer
=> f’(x) = { x3 -3x3 + 6x2}/x6
Þ f’(x) = {-2x3 + 6x2}/x6
Þ f’(x) = x2.{-2x1 + 6}/x6
Þ f’(x) = {-2x1 + 6}/x6-2
Þ f’(x) = (-2x1 + 6)/x4
So the derivative isf’(x) = (-2x1 + 6)/x4