After defining the limit and calculating a few, we introduced the limit laws. Today we do the same for the derivative. We calculate a few and introduce laws which allow us to computer more. The Power Rule shows us how to compute derivatives of polynomials, and we can also find directly the derivative of sine and cosine.

1. Section 2.3
Basic Differentiation Rules
V63.0121.027, Calculus I
September 24, 2009
Announcements
Quiz next week (up to Section 2.1)
OH today 3-4
See website for up-to-date events.
. . . . . .

2. Outline
Recall
Derivatives so far
Derivatives of power functions by hand
The Power Rule
Derivatives of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Multiple Rule
Derivatives of sine and cosine
. . . . . .

3. Derivative
. . . . . .

4. Recall: the derivative
Definition
Let f be a function and a a point in the domain of f. If the limit
f(a + h) − f(a) f(x) − f(a)
f′ (a) = lim = lim
h→0 h x→a x−a
exists, the function is said to be differentiable at a and f′ (a) is the
derivative of f at a.
The derivative …
…measures the slope of the line through (a, f(a)) tangent to
the curve y = f(x);
…represents the instantaneous rate of change of f at a
…produces the best possible linear approximation to f near
a.
. . . . . .

5. Notation
Newtonian notation Leibnizian notation
dy d df
f′ (x) y′ (x) y′ f(x)
dx dx dx
. . . . . .

6. Link between the notations
f(x + ∆x) − f(x) ∆y dy
f′ (x) = lim = lim =
∆x→0 ∆x ∆x→0 ∆x dx
dy
Leibniz thought of as a quotient of “infinitesimals”
dx
dy
We think of as representing a limit of (finite) difference
dx
quotients, not as an actual fraction itself.
The notation suggests things which are true even though they
don’t follow from the notation per se
. . . . . .

7. Outline
Recall
Derivatives so far
Derivatives of power functions by hand
The Power Rule
Derivatives of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Multiple Rule
Derivatives of sine and cosine
. . . . . .

8. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
. . . . . .

9. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x)
f′ (x) = lim
h→0 h
. . . . . .

10. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2 − x2
f′ (x) = lim = lim
h→0 h h→0 h
. . . . . .

11. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2 − x2
f′ (x) = lim = lim
h→0 h h→0 h
2
x2 x2
+ 2xh + h −
= lim
h→0 h
. . . . . .

12. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2 − x2
f′ (x) = lim = lim
h→0 h h→0 h
2 2
x2 x2
+ 2xh + h −
2xh + h¡
¡
= lim = lim
h→0 h h→0 h
¡
. . . . . .

13. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2 − x2
f′ (x) = lim = lim
h→0 h h→0 h
2 2
x2 x2
+ 2xh + h −
2xh + h¡
¡
= lim = lim
h→0 h h→0 h
¡
= lim (2x + h) = 2x.
h→0
So f′ (x) = 2x.
. . . . . .

14. The second derivative
If f is a function, so is f′ , and we can seek its derivative.
f′′ = (f′ )′
It measures the rate of change of the rate of change!
. . . . . .

15. The second derivative
If f is a function, so is f′ , and we can seek its derivative.
f′′ = (f′ )′
It measures the rate of change of the rate of change! Leibnizian
notation:
d2 y d2 d2 f
f(x)
dx2 dx2 dx2
. . . . . .

16. The squaring function and its derivatives
y
.
. x
.
. . . . . .

17. The squaring function and its derivatives
y
.
. f
.
x
.
. . . . . .

18. The squaring function and its derivatives
y
.
.′
f
f increasing =⇒ f′ ≥ 0
f
. f decreasing =⇒ f′ ≤ 0
. x
. horizontal tangent at 0
=⇒ f′ (0) = 0
. . . . . .

19. The squaring function and its derivatives
y
.
.′
f
.′′
f f increasing =⇒ f′ ≥ 0
f
. f decreasing =⇒ f′ ≤ 0
. x
. horizontal tangent at 0
=⇒ f′ (0) = 0
. . . . . .

20. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the definition of derivative to find f′ (x).
. . . . . .

21. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)3 − x3
f′ (x) = lim = lim
h→0 h h→0 h
. . . . . .

22. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)3 − x3
f′ (x) = lim = lim
h→0 h h→0 h
2 3
x3 2 x3
+ 3x h + 3xh + h −
= lim
h→0 h
. . . . . .

23. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)3 − x3
f′ (x) = lim = lim
h→0 h h→0 h
1 2
x3 2 2 3
x3 3x2 h ¡
!
2 !
¡
3
+ 3x h + 3xh + h −
¡ + 3xh + h
= lim = lim
h→0 h h→0 h
¡
. . . . . .

24. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)3 − x3
f′ (x) = lim = lim
h→0 h h→0 h
1 2
x3 2 2 3
x3 3x2 h ¡
!
2 !
¡
3
+ 3x h + 3xh + h −
¡ + 3xh + h
= lim = lim
h→0 h h→0 h
¡
( )
= lim 3x2 + 3xh + h2 = 3x2 .
h→0
So f′ (x) = 3x2 .
. . . . . .

25. The cubing function and its derivatives
y
.
. x
.
. . . . . .

26. The cubing function and its derivatives
y
.
f
.
. x
.
. . . . . .

27. The cubing function and its derivatives
Notice that f is
y
.
f
. ′ increasing, and f′ > 0
except f′ (0) = 0
f
.
. x
.
. . . . . .

28. The cubing function and its derivatives
Notice that f is
y
.
f
. ′′
f
. ′ increasing, and f′ > 0
except f′ (0) = 0
f
.
. x
.
. . . . . .

29. The cubing function and its derivatives
Notice that f is
y
.
f
. ′′
f
. ′ increasing, and f′ > 0
except f′ (0) = 0
Notice also that the
f
. tangent line to the graph
. x
. of f at (0, 0) crosses the
graph (contrary to a
popular “definition” of
the tangent line)
. . . . . .

30. Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find
f′ (x).
. . . . . .

31. Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find
f′ (x).
Solution
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
. . . . . .

32. Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find
f′ (x).
Solution
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
. . . . . .

33. Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find
f′ (x).
Solution
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(¡ + h) − ¡
x x
= lim (√ √ )
h→0 h x+h+ x
. . . . . .

34. Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find
f′ (x).
Solution
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(¡ + h) − ¡
x x h
¡
= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h→0 h
¡ x+h+ x
. . . . . .

35. Derivative of the square root function
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find
f′ (x).
Solution
√ √
′ f(x + h) − f(x) x+h− x
f (x) = lim = lim
h→0 h h→0 h
√ √ √ √
x+h− x x+h+ x
= lim ·√ √
h→0 h x+h+ x
(¡ + h) − ¡
x x h
¡
= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h→0 h
¡ x+h+ x
1
= √
2 x
√
So f′ (x) = x = 1 x−1/2 .
2 . . . . . .

36. The square root function and its derivatives
y
.
. x
.
. . . . . .

37. The square root function and its derivatives
y
.
f
.
. x
.
. . . . . .

38. The square root function and its derivatives
y
.
Here lim f′ (x) = ∞ and
f
. x→0+
f is not differentiable at 0
. .′
f
x
.
. . . . . .

39. The square root function and its derivatives
y
.
Here lim f′ (x) = ∞ and
f
. x→0+
f is not differentiable at 0
. .′
f
x
. Notice also
lim f′ (x) = 0
x→∞
. . . . . .

40. Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find
f′ (x).
. . . . . .

41. Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find
f′ (x).
Solution
f(x + h) − f(x) (x + h)1/3 − x1/3
f′ (x) = lim = lim
h→0 h h→0 h
. . . . . .

42. Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find
f′ (x).
Solution
f(x + h) − f(x) (x + h)1/3 − x1/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
. . . . . .

43. Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find
f′ (x).
Solution
f(x + h) − f(x) (x + h)1/3 − x1/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(¡ + h) − ¡
x x
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)
. . . . . .

44. Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find
f′ (x).
Solution
f(x + h) − f(x) (x + h)1/3 − x1/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(¡ + h) − ¡
x x
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)
h
¡
= lim ( )
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
¡
. . . . . .

45. Derivative of the cube root function
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find
f′ (x).
Solution
f(x + h) − f(x) (x + h)1/3 − x1/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
= lim ·
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
(¡ + h) − ¡
x x
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)
h
¡ 1
= lim ( )=
h→0 h (x + h)2/3 + (x + h)1/3 x1/3 + x2/3
¡ 3x2/3
So f′ (x) = 1 x−2/3 .
3 . . . . . .

46. The cube root function and its derivatives
y
.
. x
.
. . . . . .

47. The cube root function and its derivatives
y
.
f
.
. x
.
. . . . . .

48. The cube root function and its derivatives
y
.
Here lim f′ (x) = ∞ and f
x→0
f
. is not differentiable at 0
. .′
f
x
.
. . . . . .

49. The cube root function and its derivatives
y
.
Here lim f′ (x) = ∞ and f
x→0
f
. is not differentiable at 0
. .′
f Notice also
x
.
lim f′ (x) = 0
x→±∞
. . . . . .

50. One more
Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).
. . . . . .

51. One more
Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
. . . . . .

52. One more
Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 ( )
= lim · (x + h)1/3 + x1/3
h→0 h
. . . . . .

53. One more
Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 ( )
= lim · (x + h)1/3 + x1/3
h→0 h
( )
1 −2/3 1 /3
= 3x 2x
. . . . . .

54. One more
Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).
Solution
f(x + h) − f(x) (x + h)2/3 − x2/3
f′ (x) = lim = lim
h→0 h h→0 h
(x + h) 1/3 − x1/3 ( )
= lim · (x + h)1/3 + x1/3
h→0 h
( )
1 −2/3
= 3x 2x 1 /3
= 2 x−1/3
3
So f′ (x) = 2 x−1/3 .
3
. . . . . .

55. The function x → x2/3 and its derivative
y
.
. x
.
. . . . . .

56. The function x → x2/3 and its derivative
y
.
f
.
. x
.
. . . . . .

57. The function x → x2/3 and its derivative
y
.
f is not differentiable at
f
. 0 and lim f′ (x) = ±∞
x→0±
′
. f
.
x
.
. . . . . .

58. The function x → x2/3 and its derivative
y
.
f is not differentiable at
f
. 0 and lim f′ (x) = ±∞
x→0±
′
. f
. Notice also
x
.
lim f′ (x) = 0
x→±∞
. . . . . .

59. Recap
y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

60. Recap
y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

61. Recap
y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

62. Recap
y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

63. Recap
y y′
x2 2x
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

64. Recap
y y′
x2 2x1
x3 3x2
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

65. Recap: The Tower of Power
y y′
x2 2x1 The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

66. Recap: The Tower of Power
y y′
x2 2x The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

67. Recap: The Tower of Power
y y′
x2 2x The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

68. Recap: The Tower of Power
y y′
x2 2x The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

69. Recap: The Tower of Power
y y′
x2 2x The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

70. Recap: The Tower of Power
y y′
x2 2x The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .

71. Recap: The Tower of Power
y y′
x2 2x The power goes down
by one in each
x3 3x2 derivative
1 −1/2
x1/2 2x
The coefficient in the
1 −2/3 derivative is the power
x1/3 3x of the original function
2 −1/3
x2/3 3x
. . . . . .

72. The Power Rule
There is mounting evidence for
Theorem (The Power Rule)
Let r be a real number and f(x) = xr . Then
f′ (x) = rxr−1
as long as the expression on the right-hand side is defined.
Perhaps the most famous rule in calculus
We will assume it as of today
We will prove it many ways for many different r.
. . . . . .

73. The other Tower of Power
. . . . . .

74. Outline
Recall
Derivatives so far
Derivatives of power functions by hand
The Power Rule
Derivatives of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Multiple Rule
Derivatives of sine and cosine
. . . . . .

75. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
. . . . . .

76. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
n
∑
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k
n copies k=0
. . . . . .

77. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
n
∑
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k
n copies k=0
The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that.
. . . . . .

78. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
n
∑
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k
n copies k=0
The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that. The coefficient of
xn−1 h is the number of ways we can choose x n − 1 times, which
is the same as the number of different hs we can pick, which is
n.
. . . . . .

79. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
n
∑
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k
n copies k=0
The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that. The coefficient of
xn−1 h is the number of ways we can choose x n − 1 times, which
is the same as the number of different hs we can pick, which is
n.
. . . . . .

80. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
n
∑
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k
n copies k=0
The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that. The coefficient of
xn−1 h is the number of ways we can choose x n − 1 times, which
is the same as the number of different hs we can pick, which is
n.
. . . . . .

81. Remember your algebra
Fact
Let n be a positive whole number. Then
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
Proof.
We have
n
∑
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h) = c k x k h n −k
n copies k=0
The coefficient of xn is 1 because we have to choose x from each
binomial, and there’s only one way to do that. The coefficient of
xn−1 h is the number of ways we can choose x n − 1 times, which
is the same as the number of different hs we can pick, which is
n.
. . . . . .

82. Pascal’s Triangle
..
1
1
. 1
.
1
. 2
. 1
.
1
. 3
. 3
. 1
.
(x + h)0 = 1
1
. 4
. 6
. 4
. 1
. (x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
1
. 5 1 1 5
. .0 .0 . 1
.
(x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3
1
. 6 1 2 1 6
. .5 .0 .5 . 1
. ... ...
. . . . . .

83. Pascal’s Triangle
..
1
1
. 1
.
1
. 2
. 1
.
1
. 3
. 3
. 1
.
(x + h)0 = 1
1
. 4
. 6
. 4
. 1
. (x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
1
. 5 1 1 5
. .0 .0 . 1
.
(x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3
1
. 6 1 2 1 6
. .5 .0 .5 . 1
. ... ...
. . . . . .

84. Pascal’s Triangle
..
1
1
. 1
.
1
. 2
. 1
.
1
. 3
. 3
. 1
.
(x + h)0 = 1
1
. 4
. 6
. 4
. 1
. (x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
1
. 5 1 1 5
. .0 .0 . 1
.
(x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3
1
. 6 1 2 1 6
. .5 .0 .5 . 1
. ... ...
. . . . . .

85. Pascal’s Triangle
..
1
1
. 1
.
1
. 2
. 1
.
1
. 3
. 3
. 1
.
(x + h)0 = 1
1
. 4
. 6
. 4
. 1
. (x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
1
. 5 1 1 5
. .0 .0 . 1
.
(x + h)3 = 1x3 + 3x2 h + 3xh2 + 1h3
1
. 6 1 2 1 6
. .5 .0 .5 . 1
. ... ...
. . . . . .

86. Theorem (The Power Rule)
Let r be a positive whole number. Then
d r
x = rxr−1
dx
. . . . . .

87. Theorem (The Power Rule)
Let r be a positive whole number. Then
d r
x = rxr−1
dx
Proof.
As we showed above,
(x + h)n = xn + nxn−1 h + (stuff with at least two hs in it)
So
(x + h)n − xn nxn−1 h + (stuff with at least two hs in it)
=
h h
= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h → 0.
. . . . . .

88. The Power Rule for constants
Theorem
Let c be a constant. Then
d
c=0
dx
. . . . . .

89. The Power Rule for constants
Theorem
d 0
Let c be a constant. Then l
.ike x = 0x−1
dx
d
c=0.
dx
(although x → 0x−1 is not defined at zero.)
. . . . . .

90. The Power Rule for constants
Theorem
d 0
Let c be a constant. Then l
.ike x = 0x−1
dx
d
c=0.
dx
(although x → 0x−1 is not defined at zero.)
Proof.
Let f(x) = c. Then
f(x + h) − f(x) c−c
= =0
h h
So f′ (x) = lim 0 = 0.
h→0
. . . . . .

91. New derivatives from old
This is where the calculus starts to get really powerful!
. . . . . .

92. Calculus
. . . . . .

93. Recall the Limit Laws
Fact
Suppose lim f(x) = L and lim g(x) = M and c is a constant. Then
x→a x→a
1. lim [f(x) + g(x)] = L + M
x→a
2. lim [f(x) − g(x)] = L − M
x→a
3. lim [cf(x)] = cL
x→a
4. . . .
. . . . . .

94. Adding functions
Theorem (The Sum Rule)
Let f and g be functions and define
(f + g)(x) = f(x) + g(x)
Then if f and g are differentiable at x, then so is f + g and
(f + g)′ (x) = f′ (x) + g′ (x).
Succinctly, (f + g)′ = f′ + g′ .
. . . . . .

95. Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
. . . . . .

96. Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
. . . . . .

97. Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
. . . . . .

98. Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
f(x + h) + g(x + h) − [f(x) + g(x)]
= lim
h→0 h
f(x + h) − f(x) g(x + h) − g(x)
= lim + lim
h→0 h h→0 h
′ ′
= f (x) + g (x)
Note the use of the Sum Rule for limits. Since the limits of the
difference quotients for for f and g exist, the limit of the sum is
the sum of the limits.
. . . . . .

99. Scaling functions
Theorem (The Constant Multiple Rule)
Let f be a function and c a constant. Define
(cf)(x) = cf(x)
Then if f is differentiable at x, so is cf and
(cf)′ (x) = c · f′ (x)
Succinctly, (cf)′ = cf′ .
. . . . . .

100. Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
. . . . . .

101. Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
cf(x + h) − cf(x)
= lim
h→0 h
. . . . . .

102. Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
cf(x + h) − cf(x)
= lim
h→0 h
f(x + h) − f(x)
= c lim
h→0 h
. . . . . .

103. Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
cf(x + h) − cf(x)
= lim
h→0 h
f(x + h) − f(x)
= c lim
h→0 h
′
= c · f (x)
. . . . . .

104. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
. . . . . .

105. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solution
d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
. . . . . .

106. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solution
d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
. . . . . .

107. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solution
d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
= 2 · 3x2 + 4x3 − 17 · 12x11
. . . . . .

108. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
Solution
d ( 3 )
2x + x4 − 17x12 + 37
dx
d ( 3) d d ( ) d
= 2x + x4 + −17x12 + (37)
dx dx dx dx
d d d
= 2 x3 + x4 − 17 x12 + 0
dx dx dx
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
. . . . . .

109. Outline
Recall
Derivatives so far
Derivatives of power functions by hand
The Power Rule
Derivatives of polynomials
The Power Rule for whole number powers
The Power Rule for constants
The Sum Rule
The Constant Multiple Rule
Derivatives of sine and cosine
. . . . . .

110. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the definition:
. . . . . .

111. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the definition:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
. . . . . .

112. Angle addition formulas
See Appendix A
.
.
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
. . . . . .

113. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the definition:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
( sin x cos h + cos x sin h) − sin x
= lim
h→0 h
. . . . . .

114. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the definition:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
( sin x cos h + cos x sin h) − sin x
= lim
h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
. . . . . .

115. Two important trigonometric limits
See Section 1.4
.
sin θ
lim . =1
θ→0 θ
s
. in θ . cos θ − 1
θ lim =0
.
θ θ→0 θ
.
. − cos θ
1 1
.
. . . . . .

116. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the definition:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
( sin x cos h + cos x sin h) − sin x
= lim
h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
= sin x · 0 + cos x · 1
. . . . . .

117. Derivatives of Sine and Cosine
Fact
d
sin x = ???
dx
Proof.
From the definition:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
( sin x cos h + cos x sin h) − sin x
= lim
h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
= sin x · 0 + cos x · 1
. . . . . .

118. Derivatives of Sine and Cosine
Fact
d
sin x = cos x
dx
Proof.
From the definition:
d sin(x + h) − sin x
sin x = lim
dx h→0 h
( sin x cos h + cos x sin h) − sin x
= lim
h→0 h
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
= sin x · 0 + cos x · 1 = cos x
. . . . . .

119. Illustration of Sine and Cosine
y
.
. x
.
.
π −2
. π 0
. .π .
π
2
. in x
s
. . . . . .

120. Illustration of Sine and Cosine
y
.
. x
.
.
π −2
. π 0
. .π .
π
2 . os x
c
. in x
s
f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
. . . . . .

121. Illustration of Sine and Cosine
y
.
. x
.
.
π −2
. π 0
. .π .
π
2 . os x
c
. in x
s
f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
what happens at the horizontal tangents of cos?
. . . . . .

122. Derivatives of Sine and Cosine
Fact
d d
sin x = cos x cos x = − sin x
dx dx
. . . . . .

123. Derivatives of Sine and Cosine
Fact
d d
sin x = cos x cos x = − sin x
dx dx
Proof.
We already did the first. The second is similar (mutatis mutandis):
d cos(x + h) − cos x
cos x = lim
dx h→0 h
. . . . . .

124. Derivatives of Sine and Cosine
Fact
d d
sin x = cos x cos x = − sin x
dx dx
Proof.
We already did the first. The second is similar (mutatis mutandis):
d cos(x + h) − cos x
cos x = lim
dx h→0 h
(cos x cos h − sin x sin h) − cos x
= lim
h→0 h
. . . . . .

125. Derivatives of Sine and Cosine
Fact
d d
sin x = cos x cos x = − sin x
dx dx
Proof.
We already did the first. The second is similar (mutatis mutandis):
d cos(x + h) − cos x
cos x = lim
dx h→0 h
(cos x cos h − sin x sin h) − cos x
= lim
h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
. . . . . .

126. Derivatives of Sine and Cosine
Fact
d d
sin x = cos x cos x = − sin x
dx dx
Proof.
We already did the first. The second is similar (mutatis mutandis):
d cos(x + h) − cos x
cos x = lim
dx h→0 h
(cos x cos h − sin x sin h) − cos x
= lim
h→0 h
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
= cos x · 0 − sin x · 1 = − sin x
. . . . . .

127. What have we learned today?
The Power Rule
. . . . . .

128. What have we learned today?
The Power Rule
The derivative of a sum is the sum of the derivatives
The derivative of a constant multiple of a function is that
constant multiple of the derivative
. . . . . .

129. What have we learned today?
The Power Rule
The derivative of a sum is the sum of the derivatives
The derivative of a constant multiple of a function is that
constant multiple of the derivative
The derivative of sine is cosine
The derivative of cosine is the opposite of sine.
. . . . . .