2. Computations of Derivatives
The algebra for computing derivatives using the limit
approach gets unmanageable fast.
3. Computations of Derivatives
The algebra for computing derivatives using the limit
approach gets unmanageable fast.
Elementary functions are built in finitely many steps
using a few basic formulas, the real numbers and
the algebraic operations +,–, *, / and function
composition
4. Computations of Derivatives
The algebra for computing derivatives using the limit
approach gets unmanageable fast.
Elementary functions are built in finitely many steps
using a few basic formulas, the real numbers and
the algebraic operations +,–, *, / and function
composition–plug in a formula into another formula.
5. Computations of Derivatives
The algebra for computing derivatives using the limit
approach gets unmanageable fast.
Elementary functions are built in finitely many steps
using a few basic formulas, the real numbers and
the algebraic operations +,–, *, / and function
composition–plug in a formula into another formula.
It’s because of their structures that elementary
functions are the ones we are able to find derivatives
easily.
6. Computations of Derivatives
The algebra for computing derivatives using the limit
approach gets unmanageable fast.
Elementary functions are built in finitely many steps
using a few basic formulas, the real numbers and
the algebraic operations +,–, *, / and function
composition–plug in a formula into another formula.
It’s because of their structures that elementary
functions are the ones we are able to find derivatives
easily.
The operation of taking derivatives is called
differentiation.
7. Computations of Derivatives
The algebra for computing derivatives using the limit
approach gets unmanageable fast.
Elementary functions are built in finitely many steps
using a few basic formulas, the real numbers and
the algebraic operations +,–, *, / and function
composition–plug in a formula into another formula.
It’s because of their structures that elementary
functions are the ones we are able to find derivatives
easily.
The operation of taking derivatives is called
differentiation. We will examine how the operation of
differentiation behaves under the above operations.
8. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation.
9. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
10. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
It’s important to point out here that the corresponding
product and quotient properties of limits do not pass
on to the differentiation operation.
11. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
It’s important to point out here that the corresponding
product and quotient properties of limits do not pass
on to the differentiation operation. The product and
quotient rules of differentiation are more complicated.
12. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
It’s important to point out here that the corresponding
product and quotient properties of limits do not pass
on to the differentiation operation. The product and
quotient rules of differentiation are more complicated.
Unless stated otherwise, we assume that the
derivatives at x exist in all the theorems below.
13. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
It’s important to point out here that the corresponding
product and quotient properties of limits do not pass
on to the differentiation operation. The product and
quotient rules of differentiation are more complicated.
Unless stated otherwise, we assume that the
derivatives at x exist in all the theorems below.
The ± and Constant–Multiple Derivative Rules
14. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
It’s important to point out here that the corresponding
product and quotient properties of limits do not pass
on to the differentiation operation. The product and
quotient rules of differentiation are more complicated.
Unless stated otherwise, we assume that the
derivatives at x exist in all the theorems below.
The ± and Constant–Multiple Derivative Rules
Let f(x) and g(x) be two functions then
15. Computations of Derivatives
The following properties of limits pass on directly to
the differentiation operation. These are the sum,
difference, and constant multiplications rules of limits:
lim (f ± g) = lim f ± lim g and that lim c(f) = c lim f
It’s important to point out here that the corresponding
product and quotient properties of limits do not pass
on to the differentiation operation. The product and
quotient rules of differentiation are more complicated.
Unless stated otherwise, we assume that the
derivatives at x exist in all the theorems below.
The ± and Constant–Multiple Derivative Rules
Let f(x) and g(x) be two functions then
i. (f(x)±g(x)) ' = f '(x)±g '(x)
ii. (cf(x)) ' = c*f '(x) where c is a constant.
17. Computations of Derivatives
To verify i, note that
lim
[f(x+h) + g(x+h)] – [f(x) + g(x)]
h →0 h
=
(f(x) + g(x)) '
18. Computations of Derivatives
To verify i, note that
lim
[f(x+h) + g(x+h)] – [f(x) + g(x)]
h →0 h
=
(f(x) + g(x)) '
lim [f(x+h) – f(x)] + [g(x+h) – g(x)]
h h →0
=
19. Computations of Derivatives
To verify i, note that
lim
[f(x+h) + g(x+h)] – [f(x) + g(x)]
h →0 h
=
(f(x) + g(x)) '
lim [f(x+h) – f(x)] + [g(x+h) – g(x)]
h h →0
=
lim [f(x+h) – f(x)] [g(x+h) – g(x)]
= +
h
h →0
h
20. Computations of Derivatives
To verify i, note that
lim
[f(x+h) + g(x+h)] – [f(x) + g(x)]
h →0 h
=
(f(x) + g(x)) '
lim [f(x+h) – f(x)] + [g(x+h) – g(x)]
h h →0
=
lim [f(x+h) – f(x)] [g(x+h) – g(x)]
= +
h
h →0
h
(the sum property of limits)
lim [f(x+h) – f(x)]
h h →0 = +
lim [g(x+h) – g(x)]
h →0
h
21. Computations of Derivatives
To verify i, note that
lim
[f(x+h) + g(x+h)] – [f(x) + g(x)]
h →0 h
=
(f(x) + g(x)) '
lim [f(x+h) – f(x)] + [g(x+h) – g(x)]
h h →0
=
lim [f(x+h) – f(x)] [g(x+h) – g(x)]
= +
h
h →0
h
(the sum property of limits)
lim [f(x+h) – f(x)]
h h →0 = +
lim [g(x+h) – g(x)]
h →0
h
= f '(x) + g '(x)
22. Computations of Derivatives
To verify i, note that
lim
[f(x+h) + g(x+h)] – [f(x) + g(x)]
h →0 h
=
(f(x) + g(x)) '
lim [f(x+h) – f(x)] + [g(x+h) – g(x)]
h h →0
=
lim [f(x+h) – f(x)] [g(x+h) – g(x)]
= +
h
h →0
h
(the sum property of limits)
lim [f(x+h) – f(x)]
h h →0 = +
lim [g(x+h) – g(x)]
h →0
h
= f '(x) + g '(x)
Your turn: Verity part ii in a similar manner.
23. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
24. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
25. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
26. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
slope = 2/3
slope = 1/3
27. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
then the slope at x = a of (f + g)(x)
slope = 2/3
slope = 1/3
28. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
y=(f + g)(x)
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
then the slope at x = a of (f + g)(x)
slope = 2/3
slope = 1/3
29. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
y=(f + g)(x)
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
then the slope at x = a of (f + g)(x)
slope = 2/3
slope = 1/3
30. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
y=(f + g)(x)
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
then the slope at x = a of (f + g)(x)
or (f + g)'(a) = 1/3 + 2/3 = 1
slope
=1/3+2/3
= 1
slope = 2/3
slope = 1/3
31. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
y=(f + g)(x)
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
then the slope at x = a of (f + g)(x)
or (f + g)'(a) = 1/3 + 2/3 = 1
slope
=1/3+2/3
= 1
slope = 2/3
slope = 1/3
Likewise c*f(x) have the slope
c*f '(x) as the slope at x = a.
32. Computations of Derivatives
Geometrically, (f + g) ' = f ' + g ' says that slope of
the sum function is the sum of the slopes at any point.
Let’s illustrate this with graphs.
y=(f + g)(x)
Let the slope at x = a of f(x) be 1/3
or f '(a) = 1/3 and that
the slope at x = a of g(x) is 2/3
or g '(a) = 2/3,
then the slope at x = a of (f + g)(x)
or (f + g)'(a) = 1/3 + 2/3 = 1
slope
=1/3+2/3
= 1
slope = 2/3
Likewise c*f(x) have the slope
c*f '(x) as the slope at x = a.
The slope of y = 2x – 1 is 2,
the slope of 3(2x – 1) = 6x – 3 is 6.
slope = 1/3
33. Computations of Derivatives
However the product or quotient rule of limits that
f
lim f
lim (f)(g) = (lim f)*(lim g) and lim
g =
lim g
bear no direct relation for computing derivatives.
34. Computations of Derivatives
However the product or quotient rule of limits that
f
lim f
lim (f)(g) = (lim f)*(lim g) and lim
g =
lim g
bear no direct relation for computing derivatives.
( ) f '
fg
!! (f * g)' ≠ f ' * g ' ≠ g '
35. Computations of Derivatives
However the product or quotient rule of limits that
f
lim f
lim (f)(g) = (lim f)*(lim g) and lim
g =
lim g
bear no direct relation for computing derivatives.
fg
!! (f * g)' ≠ f ' * g ' ( )
≠ g '
f '
The Product and Quotient Rules of Derivatives
36. Computations of Derivatives
However the product or quotient rule of limits that
f
lim f
lim (f)(g) = (lim f)*(lim g) and lim
g =
lim g
bear no direct relation for computing derivatives.
fg
!! (f * g)' ≠ f ' * g ' ( )
≠
f '
g '
The Product and Quotient Rules of Derivatives
Write f for f(x) and g for g(x) then
(fg)' = f'g + fg'
37. Computations of Derivatives
However the product or quotient rule of limits that
f
lim f
lim (f)(g) = (lim f)*(lim g) and lim
g =
lim g
bear no direct relation for computing derivatives.
fg
!! (f * g)' ≠ f ' * g ' ( )
≠
f '
g '
The Product and Quotient Rules of Derivatives
Write f for f(x) and g for g(x) then
(fg)' = f'g + fg'
gf ' – fg'
g2
f
( g)' =
38. Computations of Derivatives
However the product or quotient rule of limits that
f
lim f
lim (f)(g) = (lim f)*(lim g) and lim
g =
lim g
bear no direct relation for computing derivatives.
fg
!! (f * g)' ≠ f ' * g ' ( )
≠
f '
g '
The Product and Quotient Rules of Derivatives
Write f for f(x) and g for g(x) then
(fg)' = f'g + fg'
f
=
gf ' – fg'
( g)' g2
Here are sites for the verifications of these rules.
http://en.wikipedia.org/wiki/Product_rule#Proof_of_the_product_rule
http://en.wikipedia.org/wiki/Quotient_rule
40. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G,
F
G
41. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
42. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
43. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
(F+ΔF)(G+ΔG) – FG
44. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
(F+ΔF)(G+ΔG) – FG
= shaded area
45. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
(F+ΔF)(G+ΔG) – FG
= shaded area
= three parts
46. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
(F+ΔF)(G+ΔG) – FG
= shaded area
= three parts
= F*ΔG F*ΔG
47. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
ΔF*G
(F+ΔF)(G+ΔG) – FG
= shaded area
= three parts
= F*ΔG+ΔF*G F*ΔG
48. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
ΔF*G
(F+ΔF)(G+ΔG) – FG
= shaded area
= three parts
= F*ΔG+ΔF*G+ΔF*ΔG. F*ΔG
ΔF*ΔG
49. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
ΔF*G
(F+ΔF)(G+ΔG) – FG
= shaded area
= three parts
= F*ΔG+ΔF*G+ΔF*ΔG. F*ΔG
ΔF*ΔG
If ΔF and ΔG are small then ΔF*ΔG area is negligible
50. Computations of Derivatives
Here is a geometric analogy of the Product Rule.
Given a rectangle of F * G, Suppose F is extended to
F + ΔF and G is extended to G + ΔG
F
G
ΔF
ΔG
The difference in the two
areas is
ΔF*G
(F+ΔF)(G+ΔG) – FG
= shaded area
= three parts
= F*ΔG+ΔF*G+ΔF*ΔG. F*ΔG
ΔF*ΔG
If ΔF and ΔG are small then ΔF*ΔG area is negligible
so(F+ΔF)(G+ΔG) – FG ≈ F*ΔG+ΔF*G
51. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
52. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
53. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
54. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
ii. If f(x) = x, the identity function, then f '(x) = (x) ' = 1.
55. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
ii. If f(x) = x, the identity function, then f '(x) = (x) ' = 1.
If we view x2 as the product of f(x)*g(x) where
f(x) = g(x) = x,
56. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
ii. If f(x) = x, the identity function, then f '(x) = (x) ' = 1.
If we view x2 as the product of f(x)*g(x) where
f(x) = g(x) = x, then
(x2)' = (fg)’
57. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
ii. If f(x) = x, the identity function, then f '(x) = (x) ' = 1.
If we view x2 as the product of f(x)*g(x) where
f(x) = g(x) = x, then
(x2)' = (fg)' = (f)'g + f(g)' by the Product Rule
58. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
ii. If f(x) = x, the identity function, then f '(x) = (x) ' = 1.
If we view x2 as the product of f(x)*g(x) where
f(x) = g(x) = x, then
(x2)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (1)x + x(1) by the fact (x)' = 1
59. Computations of Derivatives
Knowing how the differentiation operation ( ) '
interacts with the algebraic operations +, –, *, and /,
enables us develop the algebra to take the derivatives
of polynomials and rational functions.
Derivatives of Polynomials and Rational Functions
We start with the observations that
i. If f(x) = c, a constant function, then f '(x) = (c) ' = 0.
ii. If f(x) = x, the identity function, then f '(x) = (x) ' = 1.
If we view x2 as the product of f(x)*g(x) where
f(x) = g(x) = x, then
(x2)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (1)x + x(1) by the fact (x)' = 1
or that (x2)'= 2x
61. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)'
We say we “pull out the
constant “ when we use the
Constant Multiple Rule.
62. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
We say we “pull out the
constant “ when we use the
Constant Multiple Rule.
63. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
(–2x2 + 2x + 1)' =
64. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
“ take derivative
term by term “
(–2x2 + 2x + 1)' = (–2x2)' + (2x)' + (1)'
65. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
“ take derivative
term by term “
(–2x2 + 2x + 1)' = (–2x2)' + (2x)' + (1)'
“pull out the constant “
= –2(x2)' + 2(x)'
66. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
(–2x2 + 2x + 1)' = (–2x2)' + (2x)' + (1)'
“pull out the constant “
= –2(x2)' + 2(x)' + 0
“derivative of constant = 0“
“ take derivative
term by term “
67. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
“ take derivative
term by term “
(–2x2 + 2x + 1)' = (–2x2)' + (2x)' + (1)'
“pull out the constant “
= –2(x2)' + 2(x)' + 0
“derivative of constant = 0“
= –2(2x) + 2(1)
68. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
“ take derivative
term by term “
(–2x2 + 2x + 1)' = (–2x2)' + (2x)' + (1)'
“pull out the constant “
= –2(x2)' + 2(x)' + 0
“derivative of constant = 0“
= –2(2x) + 2(1)
= –4x + 2
69. Computations of Derivatives
Because (x2)' = 2x, therefore the derivative of –2x2 is
(–2x2)' = –2(x2)' = –2(2x) = –4x.
To calculate
“ take derivative
term by term “
(–2x2 + 2x + 1)' = (–2x2)' + (2x)' + (1)'
“pull out the constant “
= –2(x2)' + 2(x)' + 0
“derivative of constant = 0“
= –2(2x) + 2(1)
= –4x + 2
This method is a lot easier than the limit method
required by the definition.
71. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
72. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)'
73. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
74. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
so (x3)'= 3x2
75. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
so (x3)'= 3x2
If we view x4 as the product of f(x)*g(x) where
f(x) = x3 and g(x) = x and apply the result above,
then we get (x4)' = 4x3.
76. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
so (x3)'= 3x2
If we view x4 as the product of f(x)*g(x) where
f(x) = x3 and g(x) = x and apply the result above,
then we get (x4)' = 4x3.
Continue in this manner (x5)' = 5x4, (x6)' = 6x5 etc..
77. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
so (x3)'= 3x2
If we view x4 as the product of f(x)*g(x) where
f(x) = x3 and g(x) = x and apply the result above,
then we get (x4)' = 4x3.
Continue in this manner (x5)' = 5x4, (x6)' = 6x5 etc..
we obtain the formula of derivatives of the monomials.
Derivatives of Monomials
(xN)' =
78. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
so (x3)'= 3x2
If we view x4 as the product of f(x)*g(x) where
f(x) = x3 and g(x) = x and apply the result above,
then we get (x4)' = 4x3.
Continue in this manner (x5)' = 5x4, (x6)' = 6x5 etc..
we obtain the formula of derivatives of the monomials.
Derivatives of Monomials
(xN)' = NxN–1, N = 0, 1, 2, ..
79. Computations of Derivatives
If we view x3 as the product of f(x)*g(x) where
f(x) = x2 and g(x) = x, then
(x3)' = (fg)' = (f)'g + f(g)' by the Product Rule
= (x2)'x + x2(x)' = (2x)x + x2(1)
so (x3)'= 3x2
If we view x4 as the product of f(x)*g(x) where
f(x) = x3 and g(x) = x and apply the result above,
then we get (x4)' = 4x3.
Continue in this manner (x5)' = 5x4, (x6)' = 6x5 etc..
we obtain the formula of derivatives of the monomials.
Derivatives of Monomials
(xN)' = NxN–1, N = 0, 1, 2, .. and that (cxN)' = cNxN–1
where c is a constant.
81. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
82. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
(–2x7– πx3 + sin(π/4)x2 + ln(8))'
83. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
(–2x7– πx3 + sin(π/4)x2 + ln(8))'
=(–2x7)' – (πx3)' + (sin(π/4)x2)' + (ln(8))'
84. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
(–2x7– πx3 + sin(π/4)x2 + ln(8))'
=(–2x7)' – (πx3)' + (sin(π/4)x2)' + (ln(8))'
= –2 (x7)' –π (x3)' + sin(π/4) (x2)' + 0
85. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
(–2x7– πx3 + sin(π/4)x2 + ln(8))'
=(–2x7)' – (πx3)' + (sin(π/4)x2)' + (ln(8))'
= –2 (x7)' –π (x3)' + sin(π/4) (x2)' + 0
= –2 (7x6) –π (3x2) + sin(π/4) 2x + 0
86. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
(–2x7– πx3 + sin(π/4)x2 + ln(8))'
=(–2x7)' – (πx3)' + (sin(π/4)x2)' + (ln(8))'
= –2 (x7)' –π (x3)' + sin(π/4) (x2)' + 0
= –2 (7x6) –π (3x2) + sin(π/4) 2x + 0
= –14x6 – 3πx2 + (√2/2)2x
87. Computations of Derivatives
In the
dxN
dx
notation,
d
dx
= NxN–1
Example A. Find the derivatives of the following
functions.
a. –2x7– πx3 + sin(π/4)x2 + ln(8)
(–2x7– πx3 + sin(π/4)x2 + ln(8))'
=(–2x7)' – (πx3)' + (sin(π/4)x2)' + (ln(8))'
= –2 (x7)' –π (x3)' + sin(π/4) (x2)' + 0
= –2 (7x6) –π (3x2) + sin(π/4) 2x + 0
= –14x6 – 3πx2 + (√2/2)2x
= –14x6 – 3πx2 + x√2
104. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
105. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
106. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
107. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
Example B.
5
Find the derivative of f(x) = √4x2
108. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
Example B.
5
Find the derivative of f(x) = √4x2
5
(√4x2)'
= [(4x2)1/5] '
109. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
Example B.
5
Find the derivative of f(x) = √4x2
5
(√4x2)'
= [(4x2)1/5] '
= [41/5x2/5] '
110. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
Example B.
5
Find the derivative of f(x) = √4x2
5
(√4x2)'
= [(4x2)1/5] '
= [41/5x2/5] '
= 41/5 [x2/5] '
111. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
Example B.
5
Find the derivative of f(x) = √4x2
5
(√4x2)'
= [(4x2)1/5] '
= [41/5x2/5] '
= 41/5 [x2/5] '
= 41/5 [ 2 x2/5 – 1]
5
112. More Computations of Derivatives
The formulas for the derivatives of monomials may
be extended to the power function f(x) = xP where P is
any nonzero real number.
Derivatives of the Power Functions
(xP)' = PxP–1 where P is a non–zero number.
Example B.
5
Find the derivative of f(x) = √4x2
5
(√4x2)'
= [(4x2)1/5] '
= [41/5x2/5] '
= 41/5 [x2/5] '
= 41/5 [ 2 x2/5 – 1]
5
(41/5)
= 2 x –3/5
5
