1. Section 2.3
Basic Differentiation Rules
V63.0121.021, Calculus I
New York University
September 30, 2010
Announcements
Last chance for extra credit on Quiz 1: Do the get-to-know you
survey and photo by October 1.
. . . . . . .
2. Announcements
Last chance for extra credit
on Quiz 1: Do the
get-to-know you survey
and photo by October 1.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 2 / 42
3. Objectives
Understand and use these
differentiation rules:
the derivative of a
constant function (zero);
the Constant Multiple
Rule;
the Sum Rule;
the Difference Rule;
the derivatives of sine
and cosine.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 3 / 42
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 4 / 42
5. Notation
Newtonian notation Leibnizian notation
dy d df
f′ (x) y′ (x) y′ f(x)
dx dx dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 5 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 6 / 42
7. Outline
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 7 / 42
8. Derivative of the squaring function
Example
Suppose f(x) = x2 . Use the definition of derivative to find f′ (x).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 8 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 8 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 8 / 42
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
+ 2xh + h −
x2 x2
= lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 8 / 42
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
+ 2xh + h −
x2 x2 2x + h¡
h 2
= lim = lim
h→0 h h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 8 / 42
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
+ 2xh + h −
x2 x2 2x + h¡
h 2
= lim = lim
h→0 h h→0 h
= lim (2x + h) = 2x.
h→0
So f′ (x) = 2x.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 8 / 42
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!
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 9 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 9 / 42
16. The squaring function and its derivatives
y
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 10 / 42
17. The squaring function and its derivatives
y
.
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 10 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 10 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 10 / 42
20. Derivative of the cubing function
Example
Suppose f(x) = x3 . Use the definition of derivative to find f′ (x).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 11 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 11 / 42
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
+ 3x h + 3xh + h −
x3 2 x3
= lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 11 / 42
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 + 3x2 h 2
+ 3xh + h − 3
x3 3x2 h ¡
!
2 !
¡
3
+ 3xh + h
= lim = lim
h→0 h h→0
h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 11 / 42
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
+ 3xh + h −
x3 + 3x2 h 2 3
x3 3x2 h ¡
!
2 !
¡
3
+ 3xh + h
= lim = lim
h→0
(
h
)
h→0
h
2 2 2
= lim 3x + 3xh + h = 3x .
h→0
So f′ (x) = 3x2 .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 11 / 42
25. The cubing function and its derivatives
y
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 12 / 42
26. The cubing function and its derivatives
y
.
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 12 / 42
27. The cubing function and its derivatives
y
. Notice that f is increasing,
.′
f and f′ 0 except f′ (0) = 0
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 12 / 42
28. The cubing function and its derivatives
y
. Notice that f is increasing,
.′′ .′
f f and f′ 0 except f′ (0) = 0
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 12 / 42
29. The cubing function and its derivatives
y
. Notice that f is increasing,
.′′ .′
f f and f′ 0 except f′ (0) = 0
Notice also that the
f
. tangent line to the graph of
. x
. f at (0, 0) crosses the
graph (contrary to a
popular “definition” of the
tangent line)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 12 / 42
30. Derivative of the square root function
.
Example
√
Suppose f(x) = x = x1/2 . Use the definition of derivative to find f′ (x).
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 13 / 42
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 13 / 42
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 13 / 42
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
(x + h) − x
¡ ¡
= lim (√ √ )
h→0 h x+h+ x
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 13 / 42
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
(x + h) − x
¡ ¡
h
= lim (√ √ ) = lim (√ √ )
h→0 h x+h+ x h→0
h x+h+ x
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 13 / 42
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
(x + h) − 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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 13 / 42
36. The square root function and its derivatives
y
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 14 / 42
37. The square root function and its derivatives
y
.
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 14 / 42
38. The square root function and its derivatives
y
.
f
. Here lim+ f′ (x) = ∞ and f
x→0
. .′
f is not differentiable at 0
x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 14 / 42
39. The square root function and its derivatives
y
.
f
. Here lim+ f′ (x) = ∞ and f
x→0
. .′
f is not differentiable at 0
x
.
Notice also lim f′ (x) = 0
x→∞
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 14 / 42
40. Derivative of the cube root function
.
Example
√
Suppose f(x) = 3
x = x1/3 . Use the definition of derivative to find f′ (x).
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 15 / 42
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 15 / 42
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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 15 / 42
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
(x + h) − x
¡ ¡
= lim ( 2/3 + (x + h)1/3 x1/3 + x2/3
)
h→0 h (x + h)
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 15 / 42
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
(x + h) − 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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 15 / 42
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
(x + h) − 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
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 15 / 42
46. The cube root function and its derivatives
y
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 16 / 42
47. The cube root function and its derivatives
y
.
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 16 / 42
48. The cube root function and its derivatives
y
.
Here lim f′ (x) = ∞ and f is
f
. x→0
not differentiable at 0
. .′
f
x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 16 / 42
49. The cube root function and its derivatives
y
.
Here lim f′ (x) = ∞ and f is
f
. x→0
not differentiable at 0
.′
f
. x
. Notice also lim f′ (x) = 0
x→±∞
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 16 / 42
50. One more
Example
Suppose f(x) = x2/3 . Use the definition of derivative to find f′ (x).
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 17 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 17 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 17 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 17 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 17 / 42
55. The function x → x2/3 and its derivative
y
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 18 / 42
56. The function x → x2/3 and its derivative
y
.
f
.
. x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 18 / 42
57. The function x → x2/3 and its derivative
y
.
f is not differentiable at 0
f
. and lim f′ (x) = ±∞
x→0±
. .′
f
x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 18 / 42
58. The function x → x2/3 and its derivative
y
.
f is not differentiable at 0
f
. and lim f′ (x) = ±∞
x→0±
. .′
f
x
. Notice also lim f′ (x) = 0
x→±∞
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 18 / 42
65. Recap: The Tower of Power
y y′
x2 2x1 The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
66. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
67. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
68. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
69. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
70. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2
x 2x
1 −2/3
x1/3 3x
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
71. Recap: The Tower of Power
y y′
x2 2x The power goes down by
x 3
3x 2 one in each derivative
1/2 1 −1/2 The coefficient in the
x 2x derivative is the power of
1 −2/3
x1/3 3x
the original function
2 −1/3
x2/3 3x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 19 / 42
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 20 / 42
73. The other Tower of Power
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 21 / 42
74. Outline
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 22 / 42
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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
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) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−k
n copies k=0
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
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) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
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) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−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. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
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) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−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. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
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) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−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. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
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) = (x + h) · (x + h) · (x + h) · · · (x + h) =
n
ck xk hn−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. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 23 / 42
86. Proving the Power Rule
.
Theorem (The Power Rule)
Let n be a positive whole number. Then
d n
x = nxn−1
dx
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 25 / 42
87. Proving the Power Rule
.
Theorem (The Power Rule)
Let n be a positive whole number. Then
d n
x = nxn−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.
. . . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 25 / 42
88. The Power Rule for constants
Theorem
Let c be a constant. Then
d
c=0
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 26 / 42
89. The Power Rule for constants
Theorem d 0
l
.ike x = 0x−1
Let c be a constant. Then dx
d
c=0.
dx
(although x → 0x−1 is not defined at zero.)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 26 / 42
90. The Power Rule for constants
Theorem d 0
l
.ike x = 0x−1
Let c be a constant. Then 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 26 / 42
92. 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. . . .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 28 / 42
93. 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′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 29 / 42
94. Proof of the Sum Rule
Proof.
Follow your nose:
(f + g)(x + h) − (f + g)(x)
(f + g)′ (x) = lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 30 / 42
95. Proof of the Sum Rule
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 30 / 42
96. Proof of the Sum Rule
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 30 / 42
97. Proof of the Sum Rule
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.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 30 / 42
98. 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′ .
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 31 / 42
99. Proof of the Constant Multiple Rule
Proof.
Again, follow your nose.
(cf)(x + h) − (cf)(x)
(cf)′ (x) = lim
h→0 h
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 32 / 42
100. Proof of the Constant Multiple Rule
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 32 / 42
101. Proof of the Constant Multiple Rule
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 32 / 42
102. Proof of the Constant Multiple Rule
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)
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 32 / 42
103. Derivatives of polynomials
Example
d ( 3 )
Find 2x + x4 − 17x12 + 37
dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 33 / 42
104. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 33 / 42
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
d 3 d 4 d 12
= 2 x + x − 17 x + 0
dx dx dx
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 33 / 42
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 3 d 4 d 12
= 2 x + x − 17 x + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 33 / 42
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 3 d 4 d 12
= 2 x + x − 17 x + 0
dx dx dx
= 2 · 3x + 4x − 17 · 12x11
2 3
= 6x2 + 4x3 − 204x11
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 33 / 42
108. Outline
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 34 / 42
109. Derivatives of Sine and Cosine
.
Fact
d
sin x = ???
dx
Proof.
From the definition:
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 35 / 42
110. 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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 35 / 42
111. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 36 / 42
112. 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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 37 / 42
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
cos h − 1 sin h
= sin x · lim + cos x · lim
h→0 h h→0 h
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 37 / 42
114. Two important trigonometric limits
See Section 1.4
.
sin θ
lim
. =1
θ→0 θ
. in θ .
s θ cos θ − 1
lim =0
.
θ θ→0 θ
.
. − cos θ
1 1
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 38 / 42
115. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 39 / 42
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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 39 / 42
117. 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
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 39 / 42
118. Illustration of Sine and Cosine
y
.
. x
.
.
π −2
. π 0
. .π .
π
2
s
. in x
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 40 / 42
119. Illustration of Sine and Cosine
y
.
. x
.
.
π −2
. π 0
. .π .
π
2 c
. os x
s
. in x
f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 40 / 42
120. Illustration of Sine and Cosine
y
.
. x
.
.
π −2
. π 0
. .π .
π
2 c
. os x
s
. in x
f(x) = sin x has horizontal tangents where f′ = cos(x) is zero.
what happens at the horizontal tangents of cos?
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 40 / 42
121. Derivatives of Sine and Cosine
.
Fact
d d
sin x = cos x cos x = − sin x
dx dx
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 41 / 42
122. 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
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 41 / 42
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
(cos x cos h − sin x sin h) − cos x
= lim
h→0 h
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 41 / 42
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
cos h − 1 sin h
= cos x · lim − sin x · lim
h→0 h h→0 h
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 41 / 42
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
= cos x · 0 − sin x · 1 = − sin x
.
. . . . . .
V63.0121.021, Calculus I (NYU) Section 2.3 Basic Differentiation Rules September 30, 2010 41 / 42