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.
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.
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.)
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
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.
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 rational
function gets more and more complicated.
(Your turn. Find the first three derivatives of
f(x) = ln(x2 – 3).)
37. 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)
38. 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.
39. 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.
40. 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.
41. 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
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. 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.
54. 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.
55. 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.
