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Basic Mathematics

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Derivative & its Rules

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Basic Mathematics

  1. 1.  GROUP LEADER : MR.ADEEL IFTIKHAR(20397)  GROUP MEMBER : MR.FAIZAN FARAZ(20274) MR.ABDUL HASEEB(20272)
  2. 2.  DEFINITION OF DERIVATIVE  DERIVATIVE NOTATIONS  DERIVATIVE RULES  DERIVATIVE OF CONSTANT WITH EXAMPLE  POWER RULE WITH EXAMPLE  PRODUCT RULE WITH EXAMPLE  QOUTIENT RULE WITH EXAMPLE
  3. 3. 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.
  4. 4. The derivative of the position of a moving object with respect to time is the object's velocity
  5. 5. These are the notations of derivative.  F’(x) by Lagrange  dy/dx by Leibniz  Dx f(x) or Dxy by Euler  Ϋ by Newton
  6. 6. 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
  7. 7.  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)
  8. 8.  Product Rule let f(x).g(x) Then their derivative would be like = f’(x).g(x) + g’(x).f(x)
  9. 9.  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
  10. 10.  let f(x) = -55 As -55 is a constant number so applying That derivative of constant function is always zero so f’(x) = 0
  11. 11.  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
  12. 12. So by derivative of constant function is always zero. f’(x) = 0
  13. 13. ID = 20272
  14. 14. 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.
  15. 15. 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.
  16. 16. 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
  17. 17. 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
  18. 18. 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
  19. 19. 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
  20. 20. 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
  21. 21. 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
  22. 22. 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
  23. 23. ID :- 20397
  24. 24. For derivation it must be kept in mind that for derivation a function must be continues. Such that Left hand limit = Right hand limit
  25. 25. 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
  26. 26. 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)
  27. 27. 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 )
  28. 28. 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)
  29. 29. 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
  30. 30. f’(x) = 3x7 + 18x4 - 2x5 - 12x2 + 5x7 – 10x5 + 12x4 – 24x2 f’(x) = 8x7 + 30x4 - 12x5 - 36x2
  31. 31. 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
  32. 32. 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
  33. 33. 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
  34. 34. THANK YOU

Derivative & its Rules

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