Some of the most famous rules of the calculus of derivatives: the power rule, the sum rule, the constant multiple rule, and the number e defined so that e^x is its own derivative!

1. Section 3.1
Derivatives of Polynomials and Exponentials
Math 1a
February 20, 2008
Announcements
Problem Sessions Sunday, Thursday, 7pm, SC 310
ALEKS due today (10% of grade).
Office hours Wednesday 2/20 2–4pm SC 323
Midterm I Friday 2/29 in class (up to §3.2)

2. Outline
Derivatives so far
Derivatives of polynomials
The power rule for whole numbers
Linear combinations
Derivatives of exponential functions
By experimentation
The natural exponential function
Final examples

3. Derivative of the squaring function
Example
Suppose f (x) = x 2 . Use the definition of derivative to find f (x).

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

5. Derivative of the cubing function
Example
Suppose f (x) = x 3 . Use the definition of derivative to find f (x).

6. Derivative of the cubing function
Example
Suppose f (x) = x 3 . Use the definition of derivative to find f (x).
Solution
f (x + h) − f (x) (x + h)3 − x 3
f (x) = lim = lim
h→0 h h→0 h
1 2
x3 +
3x 2 h + 3xh2 + h3 x3
−
3x 2 h 2
¡ + 3xh¡ + h¡
! 3
!
= lim = lim
h→0 h h→0 h
¡
= lim 3x 2 + 3xh + h2 = 3x 2 .
h→0
So f (x) = 2x.

7. Derivative of the square root function
Example
√
Suppose f (x) = x = x 1/2 . Use the definition of derivative to find
f (x).

8. Derivative of the square root function
Example
√
Suppose f (x) = x = x 1/2 . 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

9. Derivative of the cube root function
Example
√
Suppose f (x) = 3
x = x 1/3 . Use the definition of derivative to find
f (x).

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

11. One more
Example
Suppose f (x) = x 2/3 . Use the definition of derivative to find f (x).

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

13. The Power Rule
There is mounting evidence for
Theorem (The Power Rule)
Let r be a real number and f (x) = x r . Then
f (x) = rx r −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 .

14. Outline
Derivatives so far
Derivatives of polynomials
The power rule for whole numbers
Linear combinations
Derivatives of exponential functions
By experimentation
The natural exponential function
Final examples

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

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

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

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

19. The Power Rule for constants
Theorem
Let c be a constant. Then
d
c=0
dx
d 0
Kind of like x = 0x −1 , although x → 0x −1 is not defined at
dx
zero.

20. The Power Rule for constants
Theorem
Let c be a constant. Then
d
c=0
dx
d 0
Kind of like x = 0x −1 , although x → 0x −1 is not defined at
dx
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

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

22. 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 .

23. 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.

24. 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) = cf (x)
Succinctly, (cf ) = cf .

25. Proof.
Again, follow your nose.
(cf )(x + h) − (cf )(x)
(cf ) (x) = lim
hto0 h
cf (x + h) − cf (x)
= lim
hto0 h
f (x + h) − f (x)
= c lim
hto0 h
= cf (x)

26. Derivatives of polynomials
Example
d
Find 2x 3 + x 4 − 17x 12 + 37
dx

27. Derivatives of polynomials
Example
d
Find 2x 3 + x 4 − 17x 12 + 37
dx
Solution
d
2x 3 + x 4 − 17x 12 + 37
dx
d d 4 d d
= 2x 3 + x + −17x 12 + (37)
dx dx dx dx
d d 4 d
= 2 x3 + x − 17 x 12 + 0
dx dx dx
= 2 · 3x 2 + 4x 3 − 17 · 12x 11
= 6x 2 + 4x 3 − 204x 11

28. Outline
Derivatives so far
Derivatives of polynomials
The power rule for whole numbers
Linear combinations
Derivatives of exponential functions
By experimentation
The natural exponential function
Final examples

29. Derivative of x → 2x
Example
Let f (x) = 2x . Use a calculator to estimate f (0).

30. Derivative of x → 2x
Example
Let f (x) = 2x . Use a calculator to estimate f (0).
Solution
We have
20+h − 20 2h − 1
f (0) = lim = lim ≈ 0.693147
h→0 h h→0 h

31. Example
d x
Use the previous fact to find 2 .
dx

32. Example
d x
Use the previous fact to find 2 .
dx
Solution
d x 2x+h − 2x 2x 2h − 2x
2 = lim = lim
dx h→0 h h→0 h
2 h−1
= lim 2x ·
h→0 h
2 h−1
= 2x lim
h→0 h
≈ (0.693147)2x

33. Example
d x
Use the previous fact to find 2 .
dx
Solution
d x 2x+h − 2x 2x 2h − 2x
2 = lim = lim
dx h→0 h h→0 h
2 h−1
= lim 2x ·
h→0 h
2 h−1
= 2x lim
h→0 h
≈ (0.693147)2x
Here we have a function whose derivative is a multiple of itself!
(Much different from a polynomial.)

34. Example
d x
Find 3 .
dx

35. Example
d x
Find 3 .
dx
Solution
d x 3x+h − 3x 3x 2h − 3x
3 = lim = lim
dx h→0 h h→0 h
3 h−1
= lim 3x ·
h→0 h
3 h−1
= 3x lim
h→0 h
≈ (1.09861)3x

36. Theorem
Let a > 1, and let f (x) = ax . Then
f (x) = f (0)f (x)

37. The natural exponential function
d x
If a = 2, a <1
dx x=0
d x
If a = 3, a >1
dx x=0
We would hope there is a number a between 2 and 3 such
d x
that a =1
dx x=0
We call this number e. Then by definition
d x
e = ex
dx

38. Example
Find
d 1 √
4x 2 + + 3 4 x + 6e x
dx x

39. Example
Find
d 1 √
4x 2 + + 3 4 x + 6e x
dx x
Solution
1 √
Remember = x −1 and 4 x = x 1/4 . So
x
dy 1 3
= 8x − 2 + x −3/4 + 6e x
dx x 4