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- 1. Higher Derivatives
- 2. Higher DerivativesLet f(x) = x2, its derivative f (x) = 2x.
- 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. 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. 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. 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. 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. 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. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.
- 10. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2
- 11. Higher DerivativesExample A. Find the first five derivatives off(x) = 2x4 – x3 – 2.f (x) = 8x3 – 3x2f (x) = 24x2 – 6x
- 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. 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. 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. 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. 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. 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. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3
- 19. Higher DerivativesExample B. Find the first three derivatives off(x) = –x2/3f (x)= –(2/3)x –1/3 = –2 3x 1/3
- 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. 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. 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. 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. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).
- 25. Higher DerivativesExample C. Find the first four derivatives off(x) = In(x).d In(x) = x–1 = 1 dx x
- 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. 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. 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. 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. 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. 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. 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. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).
- 34. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)
- 35. Higher DerivativesExample C.Find the first four derivatives of f(x) = sin(x).f (x) = cos (x)f (x) = –sin(x)
- 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. 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. 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. 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. 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. 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. Higher DerivativesIf f(x) has of an exponential factor eu(x) so thatf(x) = eu(x) v(x) for some v(x),
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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.

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