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3.1 higher derivatives

The document discusses higher order derivatives. It defines the nth derivative of a function f(x) as f(n)(x). The first example finds the first five derivatives of f(x)=2x^4 - x^3 - 2. The second example finds the first three derivatives of f(x)=-x^2/3. The third example finds the first four derivatives of f(x)=ln(x) and discusses how derivatives of rational functions become more complicated with higher orders. It also provides examples of finding derivatives of other functions like sin(x).

Higher Derivatives
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x.
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2.
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x).
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0.
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0. This is called the 3rd derivative of f(x) and it is denoted as f '''(x) or as f(3)(x).
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0. This is called the 3rd derivative of f(x) and it is denoted as f '''(x) or as f(3)(x). Continuing in this manner we define the nth derivative of f(x) as f(n)(x) for n = 1, 2, 3..
Higher Derivatives Let f(x) = x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0. This is called the 3rd derivative of f(x) and it is denoted as f '''(x) or as f(3)(x). Continuing in this manner we define the nth derivative of f(x) as f(n)(x) for n = 1, 2, 3.. We write these as dxn dnf in the d/dx-notation.
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2.
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1).
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1). Continue taking derivatives, we see that the Nth derivative of a degree N polynomial is a non–zero constant
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1). Continue taking derivatives, we see that the Nth derivative of a degree N polynomial is a non–zero constant and that its (N+1)th derivative is 0.
Higher Derivatives Example A. Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1). Continue taking derivatives, we see that the Nth derivative of a degree N polynomial is a non–zero constant and that its (N+1)th derivative is 0. If P(x) = ax N + … then P(N)(x) = c ≠ 0, a non–zero constant and its (N+1)th derivative PN+1(x) = 0.
Higher Derivatives Example B. Find the first three derivatives of f(x) = –x2/3
Higher Derivatives Example B. Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3
Higher Derivatives Example B. Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3
Higher Derivatives Example B. Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3 –8 f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3
Higher Derivatives Example B. Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3 –8 f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3 The successive derivatives of the power functions f(x) = xp in general get more complicated . (unless p = 0, 1, 2, 3… i.e. f(x) is a polynomial).
Higher Derivatives Example B. Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3 –8 f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3 The successive derivatives of the power functions f(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 of f(x) = √2 – 3x and note the changes in the exponents in the successive derivatives.)
Higher Derivatives Example C. Find the first four derivatives of f(x) = In(x).
Higher Derivatives Example C. Find the first four derivatives of f(x) = In(x). d In(x) = x–1 = 1 dx x
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
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) dx 3 = 2x–3
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
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.
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.
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 rational function gets more and more complicated.
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 rational function gets more and more complicated. (Your turn. Find the first three derivatives of f(x) = ln(x2 – 3).)
Higher Derivatives Example C. Find the first four derivatives of f(x) = sin(x).
Higher Derivatives Example C. Find the first four derivatives of f(x) = sin(x). f '(x) = cos (x)
Higher Derivatives Example C. Find the first four derivatives of f(x) = sin(x). f '(x) = cos (x) f ''(x) = –sin(x)
Higher Derivatives Example C. Find the first four derivatives of f(x) = sin(x). f '(x) = cos (x) f ''(x) = –sin(x) f(3)(x) = –cos (x)
Higher Derivatives Example 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)
Higher Derivatives Example 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 the successive derivatives of the sine function are cyclical.
Higher Derivatives Example 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 the successive derivatives of the sine function –s = s(2) = s(6) = s(10) .. are cyclical.
Higher Derivatives Example 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 the successive derivatives of the sine function –s = s(2) = s(6) = s(10) .. are cyclical. The successive derivatives of the cosine are also cyclical.
Higher Derivatives Example 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 the successive derivatives of the sine function –s = s(2) = s(6) = s(10) .. are cyclical. The successive derivatives of the cosine are also cyclical. However the successive derivatives for the other trig–functions get more complicated because they are fractions of sine and cosine
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x),
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(x) = [eu(x) v(x)] ' =
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(x) = [eu(x) v(x)] ' = [eu(x)]' v(x) + eu(x) [v(x)] '
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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)] '
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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)
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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)]
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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)].
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)].
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)]. “#(x)” means “some function in x”
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)]. This implies that all higher derivatives also are of this form – with the factor eu(x).
Higher Derivatives If f(x) has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)]. This implies that all higher derivatives also are of this form – with the factor eu(x). (Your turn. Verify that both the 1st and 2nd derivatives of f(x) = cos(x) * esin(x) has the factor esin(x). In fact, its easier to do its 2nd derivative by factoring the 1st derivative and using the Product Rule.
Higher Derivatives 2. The other factor [u'(x)v(x) + v'(x)] is a useful form due to the presence of v(x) with the addition of v', its derivative.
Higher Derivatives 2. The other factor [u'(x)v(x) + v'(x)] is a useful form due to the presence of v(x) with the addition of v', its derivative.
Higher Derivatives 2. The other factor [u'(x)v(x) + v'(x)] is a useful form due to the presence of v(x) with the addition of v', its derivative. The above form will be used in solving differential equations.

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In this document
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Introduction and definitions of higher derivatives. It provides examples distinguishing first, second, and third derivatives.

Calculation of the first five derivatives of a polynomial function f(x) = 2x⁴ - x³ - 2.

Finding derivatives of the power function f(x) = -x²/³, noting the complexity of successive derivatives.

Finding the first four derivatives of f(x) = ln(x), and applying the Log-Chain rule with polynomial functions.

Calculating the first four derivatives of f(x) = sin(x), highlighting cyclical patterns in derivatives.

Explains derivatives involving exponential factors. Derivative forms using product rule for functions containing e^{u(x)}.

3.1 higher derivatives

  • 1.
  • 2.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x.
  • 3.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2.
  • 4.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x).
  • 5.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0.
  • 6.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0. This is called the 3rd derivative of f(x) and it is denoted as f '''(x) or as f(3)(x).
  • 7.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0. This is called the 3rd derivative of f(x) and it is denoted as f '''(x) or as f(3)(x). Continuing in this manner we define the nth derivative of f(x) as f(n)(x) for n = 1, 2, 3..
  • 8.
    Higher Derivatives Let f(x)= x2, its derivative f '(x) = 2x. If we take the derivative of the derivative again we have that [f '(x)] ' = 2. This is called the 2nd derivative of f(x) and it is denoted as f ''(x) or as f(2)(x). If we take the derivative of the 2nd derivative again we have that [f ''(x)] ' = (2) ' = 0. This is called the 3rd derivative of f(x) and it is denoted as f '''(x) or as f(3)(x). Continuing in this manner we define the nth derivative 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 Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2.
  • 10.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2
  • 11.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x
  • 12.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6
  • 13.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0
  • 14.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1).
  • 15.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1). Continue taking derivatives, we see that the Nth derivative of a degree N polynomial is a non–zero constant
  • 16.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1). Continue taking derivatives, we see that the Nth derivative of a degree N polynomial is a non–zero constant and that its (N+1)th derivative is 0.
  • 17.
    Higher Derivatives Example A.Find the first five derivatives of f(x) = 2x4 – x3 – 2. f '(x) = 8x3 – 3x2 f ''(x) = 24x2 – 6x f(3)(x) = 48x – 6 f(4)(x) = 48 and f(5)(x) = 0 The derivative of a degree N polynomial is another polynomial of degree (N – 1). Continue taking derivatives, we see that the Nth derivative of a degree N polynomial is a non–zero constant and that its (N+1)th derivative is 0. If P(x) = ax N + … then P(N)(x) = c ≠ 0, a non–zero constant and its (N+1)th derivative PN+1(x) = 0.
  • 18.
    Higher Derivatives Example B.Find the first three derivatives of f(x) = –x2/3
  • 19.
    Higher Derivatives Example B.Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3
  • 20.
    Higher Derivatives Example B.Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3
  • 21.
    Higher Derivatives Example B.Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3 –8 f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3
  • 22.
    Higher Derivatives Example B.Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3 –8 f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3 The successive derivatives of the power functions f(x) = xp in general get more complicated . (unless p = 0, 1, 2, 3… i.e. f(x) is a polynomial).
  • 23.
    Higher Derivatives Example B.Find the first three derivatives of f(x) = –x2/3 f '(x)= –(2/3)x –1/3 = –2 3x 1/3 –4/3 = 2 f ''(x)= –(2/3)(–1/3)x 9x4/3 –8 f (3) (x)= –(2/3)(–1/3)(–4/3)x –7/3 = 27x7/3 The successive derivatives of the power functions f(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 of f(x) = √2 – 3x and note the changes in the exponents in the successive derivatives.)
  • 24.
    Higher Derivatives Example C.Find the first four derivatives of f(x) = In(x).
  • 25.
    Higher Derivatives Example C.Find the first four derivatives of f(x) = In(x). d In(x) = x–1 = 1 dx x
  • 26.
    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
  • 27.
    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) dx 3 = 2x–3
  • 28.
    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
  • 29.
    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.
  • 30.
    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.
  • 31.
    Higher Derivatives ExampleC. 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 rational function gets more and more complicated.
  • 32.
    Higher Derivatives ExampleC. 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 rational function gets more and more complicated. (Your turn. Find the first three derivatives of f(x) = ln(x2 – 3).)
  • 33.
    Higher Derivatives Example C. Findthe first four derivatives of f(x) = sin(x).
  • 34.
    Higher Derivatives Example C. Findthe first four derivatives of f(x) = sin(x). f '(x) = cos (x)
  • 35.
    Higher Derivatives Example C. Findthe first four derivatives of f(x) = sin(x). f '(x) = cos (x) f ''(x) = –sin(x)
  • 36.
    Higher Derivatives Example C. Findthe first four derivatives of f(x) = sin(x). f '(x) = cos (x) f ''(x) = –sin(x) f(3)(x) = –cos (x)
  • 37.
    Higher Derivatives Example C. Findthe 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 Derivatives Example C. Findthe 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 the successive derivatives of the sine function are cyclical.
  • 39.
    Higher Derivatives Example C. Findthe 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 the successive derivatives of the sine function –s = s(2) = s(6) = s(10) .. are cyclical.
  • 40.
    Higher Derivatives Example C. Findthe 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 the successive derivatives of the sine function –s = s(2) = s(6) = s(10) .. are cyclical. The successive derivatives of the cosine are also cyclical.
  • 41.
    Higher Derivatives Example C. Findthe 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 the successive derivatives of the sine function –s = s(2) = s(6) = s(10) .. are cyclical. The successive derivatives of the cosine are also cyclical. However the successive derivatives for the other trig–functions get more complicated because they are fractions of sine and cosine
  • 42.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x),
  • 43.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(x) = [eu(x) v(x)] ' =
  • 44.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(x) = [eu(x) v(x)] ' = [eu(x)]' v(x) + eu(x) [v(x)] '
  • 45.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)].
  • 50.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)]. “#(x)” means “some function in x”
  • 51.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)]. This implies that all higher derivatives also are of this form – with the factor eu(x).
  • 52.
    Higher Derivatives If f(x)has of an exponential factor eu(x) so that f(x) = eu(x) v(x) for some v(x), by the Product Rule f '(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 of the form eu(x) [#(x)]. This implies that all higher derivatives also are of this form – with the factor eu(x). (Your turn. Verify that both the 1st and 2nd derivatives of f(x) = cos(x) * esin(x) has the factor esin(x). In fact, its easier to do its 2nd derivative by factoring the 1st derivative and using the Product Rule.
  • 53.
    Higher Derivatives 2. Theother factor [u'(x)v(x) + v'(x)] is a useful form due to the presence of v(x) with the addition of v', its derivative.
  • 54.
    Higher Derivatives 2. Theother factor [u'(x)v(x) + v'(x)] is a useful form due to the presence of v(x) with the addition of v', its derivative.
  • 55.
    Higher Derivatives 2. Theother factor [u'(x)v(x) + v'(x)] is a useful form due to the presence of v(x) with the addition of v', its derivative. The above form will be used in solving differential equations.