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# 3.1 higher derivatives

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### 3.1 higher derivatives

1. 1. Higher Derivatives
2. 2. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.
3. 3. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.If we take the derivative of the derivative againwe have that [f (x)] = 2.
4. 4. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.If we take the derivative of the derivative againwe have that [f (x)] = 2.This is called the 2nd derivative of f(x) andit is denoted as f (x) or as f(2)(x).
5. 5. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.If we take the derivative of the derivative againwe have that [f (x)] = 2.This is called the 2nd derivative of f(x) andit is denoted as f (x) or as f(2)(x).If we take the derivative of the 2nd derivativeagain we have that [f (x)] = (2) = 0.
6. 6. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.If we take the derivative of the derivative againwe have that [f (x)] = 2.This is called the 2nd derivative of f(x) andit is denoted as f (x) or as f(2)(x).If we take the derivative of the 2nd derivativeagain we have that [f (x)] = (2) = 0.This is called the 3rd derivative of f(x) andit is denoted as f (x) or as f(3)(x).
7. 7. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.If we take the derivative of the derivative againwe have that [f (x)] = 2.This is called the 2nd derivative of f(x) andit is denoted as f (x) or as f(2)(x).If we take the derivative of the 2nd derivativeagain we have that [f (x)] = (2) = 0.This is called the 3rd derivative of f(x) andit is denoted as f (x) or as f(3)(x).Continuing in this manner we define the nthderivative of f(x) as f(n)(x) for n = 1, 2, 3..
8. 8. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.If we take the derivative of the derivative againwe have that [f (x)] = 2.This is called the 2nd derivative of f(x) andit is denoted as f (x) or as f(2)(x).If we take the derivative of the 2nd derivativeagain we have that [f (x)] = (2) = 0.This is called the 3rd derivative of f(x) andit is denoted as f (x) or as f(3)(x).Continuing in this manner we define the nthderivative of f(x) as f(n)(x) for n = 1, 2, 3..We write these as dxn dnf in the d/dx-notation.
9. 9. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.
10. 10. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2
11. 11. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6x
12. 12. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6xf(3)(x) = 48x – 6
13. 13. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6xf(3)(x) = 48x – 6f(4)(x) = 48 and f(5)(x) = 0
14. 14. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6xf(3)(x) = 48x – 6f(4)(x) = 48 and f(5)(x) = 0The derivative of a degree N polynomial is anotherpolynomial of degree (N – 1).
15. 15. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6xf(3)(x) = 48x – 6f(4)(x) = 48 and f(5)(x) = 0The derivative of a degree N polynomial is anotherpolynomial of degree (N – 1). Continue takingderivatives, we see that the Nth derivative of adegree N polynomial is a non–zero constant
16. 16. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6xf(3)(x) = 48x – 6f(4)(x) = 48 and f(5)(x) = 0The derivative of a degree N polynomial is anotherpolynomial of degree (N – 1). Continue takingderivatives, we see that the Nth derivative of adegree N polynomial is a non–zero constant and thatits (N+1)th derivative is 0.
17. 17. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6xf(3)(x) = 48x – 6f(4)(x) = 48 and f(5)(x) = 0The derivative of a degree N polynomial is anotherpolynomial of degree (N – 1). Continue takingderivatives, we see that the Nth derivative of adegree N polynomial is a non–zero constant and thatits (N+1)th derivative is 0.If P(x) = ax N + … then P(N)(x) = c ≠ 0, a non–zeroconstant and its (N+1)th derivative PN+1(x) = 0.
18. 18. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3
19. 19. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3f (x)= –(2/3)x –1/3 = –2 3x 1/3
20. 20. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3f (x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2f (x)= –(2/3)(–1/3)x 9x4/3
21. 21. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3f (x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2f (x)= –(2/3)(–1/3)x 9x4/3 –8f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3
22. 22. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3f (x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2f (x)= –(2/3)(–1/3)x 9x4/3 –8f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3The successive derivatives of the power functionsf(x) = xp in general get more complicated .(unless p = 0, 1, 2, 3… i.e. f(x) is a polynomial).
23. 23. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3f (x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2f (x)= –(2/3)(–1/3)x 9x4/3 –8f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3The successive derivatives of the power functionsf(x) = xp in general get more complicated .(unless p = 0, 1, 2, 3… i.e. f(x) is a polynomial).(Your turn. Find the first three derivatives off(x) = √2 – 3x and note the changes in theexponents in the successive derivatives.)
24. 24. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).
25. 25. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 dx x
26. 26. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2
27. 27. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2 d3 ln(x) dx 3 = 2x–3
28. 28. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2 d3 ln(x) d4 ln(x) dx 3 = 2x–3 dx 4 = –6x–4
29. 29. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2 d3 ln(x) d4 ln(x) dx 3 = 2x–3 dx 4 = –6x–4By the Log–Chain Rule we see that d In(P(x)) = P(x) dx P(x)is a rational function if P(x) is polynomial.
30. 30. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2 d3 ln(x) d4 ln(x) dx 3 = 2x–3 dx 4 = –6x–4By the Log–Chain Rule we see that d In(P(x)) = P(x) dx P(x)is a rational function if P(x) is polynomial.This is true because P(x) is another polynomial.
31. 31. Higher Derivatives Example C. Find the first four derivatives of f(x) = In(x). d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2 d3 ln(x) d4 ln(x) dx 3 = 2x–3 dx 4 = –6x–4 By the Log–Chain Rule we see that d In(P(x)) = P(x) dx P(x) is a rational function if P(x) is polynomial.This is true because P(x) is another polynomial.Again the successive derivatives of a rationalfunction gets more and more complicated.
32. 32. Higher Derivatives Example C. Find the first four derivatives of f(x) = In(x). d In(x) = x–1 = 1 d2 ln(x) = –x–2 = –1 dx x dx 2 x2 d3 ln(x) d4 ln(x) dx 3 = 2x–3 dx 4 = –6x–4 By the Log–Chain Rule we see that d In(P(x)) = P(x) dx P(x) is a rational function if P(x) is polynomial.This is true because P(x) is another polynomial.Again the successive derivatives of a rationalfunction gets more and more complicated.(Your turn. Find the first three derivatives off(x) = ln(x2 – 3).)
33. 33. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).
34. 34. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)
35. 35. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)f (x) = –sin(x)
36. 36. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)f (x) = –sin(x)f(3)(x) = –cos (x)
37. 37. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)f (x) = –sin(x)f(3)(x) = –cos (x)f(4)(x) = sin(x)
38. 38. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)f (x) = –sin(x)f(3)(x) = –cos (x)f(4)(x) = sin(x)Note that thesuccessive derivativesof the sine functionare cyclical.
39. 39. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x) s = s(4) = s(8) = s(12) .. s=sin(x)f (x) = –sin(x) c=cos (x)f(3)(x) = –cos (x) –c = s(3) = s(7) = s(11) .. c = s(1) = s(5) = s(9) ..f(4)(x) = sin(x)Note that thesuccessive derivativesof the sine function –s = s(2) = s(6) = s(10) ..are cyclical.
40. 40. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x) s = s(4) = s(8) = s(12) .. s=sin(x)f (x) = –sin(x) c=cos (x)f(3)(x) = –cos (x) –c = s(3) = s(7) = s(11) .. c = s(1) = s(5) = s(9) ..f(4)(x) = sin(x)Note that thesuccessive derivativesof the sine function –s = s(2) = s(6) = s(10) ..are cyclical. The successive derivatives of the cosineare also cyclical.
41. 41. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x) s = s(4) = s(8) = s(12) .. s=sin(x)f (x) = –sin(x) c=cos (x)f(3)(x) = –cos (x) –c = s(3) = s(7) = s(11) .. c = s(1) = s(5) = s(9) ..f(4)(x) = sin(x)Note that thesuccessive derivativesof the sine function –s = s(2) = s(6) = s(10) ..are cyclical. The successive derivatives of the cosineare also cyclical. However the successive derivativesfor the other trig–functions get more complicatedbecause they are fractions of sine and cosine
42. 42. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x),
43. 43. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] =
44. 44. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)]
45. 45. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] by the Chain Rule= [u(x)] eu(x) v(x) + eu(x) [v(x)]
46. 46. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] by the Chain Rule= [u(x)] eu(x) v(x) + eu(x) [v(x)] pull out eu(x)
47. 47. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] by the Chain Rule= [u(x)] eu(x) v(x) + eu(x) [v(x)] pull out eu(x)= eu(x) [u(x)v(x) + v(x)]
48. 48. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] by the Chain Rule= [u(x)] eu(x) v(x) + eu(x) [v(x)] = eu(x) [u(x)v(x) + v(x)]Two observations about the derivative of eu(x) [v(x)].
49. 49. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] = [u(x)] eu(x) v(x) + eu(x) [v(x)] by the Chain Rule= eu(x) [u(x)v(x) + v(x)]Two observations about the derivative of eu(x) [v(x)].1. The derivative of eu(x) v(x) is another function ofthe form eu(x) [#(x)].
50. 50. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] = [u(x)] eu(x) v(x) + eu(x) [v(x)] by the Chain Rule= eu(x) [u(x)v(x) + v(x)]Two observations about the derivative of eu(x) [v(x)].1. The derivative of eu(x) v(x) is another function ofthe form eu(x) [#(x)]. “#(x)” means “some function in x”
51. 51. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] = [u(x)] eu(x) v(x) + eu(x) [v(x)] by the Chain Rule= eu(x) [u(x)v(x) + v(x)]Two observations about the derivative of eu(x) [v(x)].1. The derivative of eu(x) v(x) is another function ofthe form eu(x) [#(x)]. This implies that all higherderivatives also are of this form – with the factor eu(x).
52. 52. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x), by the Product Rulef (x) = [eu(x) v(x)] = [eu(x)] v(x) + eu(x) [v(x)] = [u(x)] eu(x) v(x) + eu(x) [v(x)] by the Chain Rule= eu(x) [u(x)v(x) + v(x)]Two observations about the derivative of eu(x) [v(x)].1. The derivative of eu(x) v(x) is another function ofthe form eu(x) [#(x)]. This implies that all higherderivatives also are of this form – with the factor eu(x).(Your turn. Verify that both the 1st and 2nd derivativesof f(x) = cos(x) * esin(x) has the factor esin(x).In fact, its easier to do its 2nd derivative by factoringthe 1st derivative and using the Product Rule.
53. 53. Higher Derivatives2. The other factor [u(x)v(x) + v(x)] is a useful formdue to the presence of v(x) with the addition of v,its derivative.
54. 54. Higher Derivatives2. The other factor [u(x)v(x) + v(x)] is a useful formdue to the presence of v(x) with the addition of v,its derivative.
55. 55. Higher Derivatives2. The other factor [u(x)v(x) + v(x)] is a useful formdue to the presence of v(x) with the addition of v,its derivative. The above form will be used in solvingdifferential equations.