Lesson 8: Derivatives of Polynomials and Exponential functions

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
IProblem Sessions Sunday, Thursday, 7pm, SC 310
IALEKS due today (10% of grade).
IOce hours Wednesday 2/20 2{4pm SC 323
IMidterm I Friday 2/29 in class (up tox3.2)

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
Supposef(x) =x
2
. Use the denition of derivative to ndf
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
2
x
2
h
= lim
h!0

x
2
+ 2xh+h
2

x
2
h
= lim
h!0
2xh+h
2
h
= lim
h!0
(2x+h) = 2x:
So f
0
(x) = 2x.

4. Derivative of the squaring function
Example
Supposef(x) =x
2
. Use the denition of derivative to ndf
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
2
x
2
h
= lim
h!0

x
2
+ 2xh+h
2

x
2
h
= lim
h!0
2xh+h
2
h
= lim
h!0
(2x+h) = 2x:
So f
0
(x) = 2x.

5. Derivative of the cubing function
Example
Supposef(x) =x
3
. Use the denition of derivative to ndf
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
3
x
3
h
= lim
h!0

x
3
+ 3x
2
h+ 3xh
2
+h
3

x
3
h
= lim
h!0
3x
2
h+ 3xh
1
2
+h
2
3
h
= lim
h!0

3x
2
+ 3xh+h
2

= 3x
2
:
So f
0
(x) = 2x.

6. Derivative of the cubing function
Example
Supposef(x) =x
3
. Use the denition of derivative to ndf
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
3
x
3
h
= lim
h!0

x
3
+ 3x
2
h+ 3xh
2
+h
3

x
3
h
= lim
h!0
3x
2
h+ 3xh
1
2
+h
2
3
h
= lim
h!0

3x
2
+ 3xh+h
2

= 3x
2
:
So f
0
(x) = 2x.

7. Derivative of the square root function
Example
Supposef(x) =
p
x=x
1=2
. Use the denition of derivative to nd
f
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
p
x+h
p
x
h
= lim
h!0
p
x+h
p
x
h

p
x+h+
p
x
p
x+h+
p
x
= lim
h!0
(x+h)x
h
p
x+h+
p
x
= lim
h!0
h
h
p
x+h+
p
x

=
1
2
p
x
So f
0
(x) =
p
x=
1
2
x
1=2
.

8. Derivative of the square root function
Example
Supposef(x) =
p
x=x
1=2
. Use the denition of derivative to nd
f
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
p
x+h
p
x
h
= lim
h!0
p
x+h
p
x
h

p
x+h+
p
x
p
x+h+
p
x
= lim
h!0
(x+h)x
h
p
x+h+
p
x
= lim
h!0
h
h
p
x+h+
p
x

=
1
2
p
x
So f
0
(x) =
p
x=
1
2
x
1=2
.

9. Derivative of the cube root function
Example
Supposef(x) =
3
p
x=x
1=3
. Use the denition of derivative to nd
f
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
1=3
x
1=3
h
= lim
h!0
(x+h)
1=3
x
1=3
h

(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3
(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3
= lim
h!0
(x+h)x
h

(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3

= lim
h!0
h
h

(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3
=
1
3x
2=3
So f
0
(x) =
1
3
x
2=3
.

10. Derivative of the cube root function
Example
Supposef(x) =
3
p
x=x
1=3
. Use the denition of derivative to nd
f
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
1=3
x
1=3
h
= lim
h!0
(x+h)
1=3
x
1=3
h

(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3
(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3
= lim
h!0
(x+h)x
h

(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3

= lim
h!0
h
h

(x+h)
2=3
+ (x+h)
1=3
x
1=3
+x
2=3
=
1
3x
2=3
So f
0
(x) =
1
3
x
2=3
.

11. One more
Example
Supposef(x) =x
2=3
. Use the denition of derivative to ndf
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
2=3
x
2=3
h
= lim
h!0
(x+h)
1=3
x
1=3
h


(x+h)
1=3
+x
1=3

=
1
3
x
2=3

2x
1=3

=
2
3
x
1=3
So f
0
(x) =
2
3
x
1=3
.

12. One more
Example
Supposef(x) =x
2=3
. Use the denition of derivative to ndf
0
(x).
Solution
f
0
(x) = lim
h!0
f(x+h)f(x)
h
= lim
h!0
(x+h)
2=3
x
2=3
h
= lim
h!0
(x+h)
1=3
x
1=3
h


(x+h)
1=3
+x
1=3

=
1
3
x
2=3

2x
1=3

=
2
3
x
1=3
So f
0
(x) =
2
3
x
1=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
0
(x) =rx
r1
as long as the expression on the right-hand side is dened.
IPerhaps the most famous rule in calculus
IWe will assume it as of today
IWe will prove it many ways for many dierentr.

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
n1
h+(stu with at least two hs in it)
Proof.
We have
(x+h)
n
= (x+h)(x+h) (x+h)
| {z }
ncopies
Each monomial is of the formckx
k
h
nk
. The coecient ofx
n
is
one because we have to choosexfrom each binomial, and there's
only one way to do that. The coecient ofx
n1
his the number of
ways we can choosex n1 times, which is the same as the
number of dierenths we can pick, which isn.

16. Remember your algebra
Fact
Let n be a positive whole number. Then
(x+h)
n
=x
n
+nx
n1
h+(stu with at least two hs in it)
Proof.
We have
(x+h)
n
= (x+h)(x+h) (x+h)
| {z }
ncopies
Each monomial is of the formckx
k
h
nk
. The coecient ofx
n
is
one because we have to choosexfrom each binomial, and there's
only one way to do that. The coecient ofx
n1
his the number of
ways we can choosex n1 times, which is the same as the
number of dierenths we can pick, which isn.

17. Theorem (The Power Rule)
Let r be a positive whole number. Then
d
dx
x
r
=rx
r1
Proof.
As we showed above,
(x+h)
n
=x
n
+nx
n1
h+ (stu with at least twohs in it)
So
(x+h)
n
x
n
h
=
nx
n1
h+ (stu with at least twohs in it)
h
=nx
n1
+ (stu with at least onehin it)
and this tends tonx
n1
ash!0.

18. Theorem (The Power Rule)
Let r be a positive whole number. Then
d
dx
x
r
=rx
r1
Proof.
As we showed above,
(x+h)
n
=x
n
+nx
n1
h+ (stu with at least twohs in it)
So
(x+h)
n
x
n
h
=
nx
n1
h+ (stu with at least twohs in it)
h
=nx
n1
+ (stu with at least onehin it)
and this tends tonx
n1
ash!0.

19. The Power Rule for constants
Theorem
Let c be a constant. Then
d
dx
c= 0
Kind of like
d
dx
x
0
= 0x
1
, althoughx7!0x
1
is not dened at
zero.
Proof.
Letf(x) =c. Then
f(x+h)f(x)
h
=
cc
h
= 0
Sof
0
(x) = lim
h!0
0 = 0.

20. The Power Rule for constants
Theorem
Let c be a constant. Then
d
dx
c= 0
Kind of like
d
dx
x
0
= 0x
1
, althoughx7!0x
1
is not dened at
zero.
Proof.
Letf(x) =c. Then
f(x+h)f(x)
h
=
cc
h
= 0
Sof
0
(x) = lim
h!0
0 = 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 dene
(f+g)(x) =f(x) +g(x)
Then if f and g are dierentiable at x, then so is f+g and
(f+g)
0
(x) =f
0
(x) +g
0
(x):
Succinctly,(f+g)
0
=f
0
+g
0
.

23. Proof.
Follow your nose:
(f+g)
0
(x) = lim
h!0
(f+g)(x+h)(f+g)(x)
h
= lim
h!0
f(x+h) +g(x+h)[f(x) +g(x)]
h
= lim
h!0
f(x+h)f(x)
h
+ lim
h!0
g(x+h)g(x)
h
=f
0
(x) +g
0
(x)
Note the use of the Sum Rule for limits. Since the limits of the
dierence quotients for forfandgexist, 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. Dene
(cf)(x) =cf(x)
Then if f is dierentiable at x, so is cf and
(cf)
0
(x) =cf
0
(x)
Succinctly,(cf)
0
=cf
0
.

25. Proof.
Again, follow your nose.
(cf)
0
(x) = lim
hto0
(cf)(x+h)(cf)(x)
h
= lim
hto0
cf(x+h)cf(x)
h
=clim
hto0
f(x+h)f(x)
h
=cf
0
(x)

26. Derivatives of polynomials
Example
Find
d
dx

2x
3
+x
4
17x
12
+ 37

Solution
d
dx

2x
3
+x
4
17x
12
+ 37

=
d
dx

2x
3

+
d
dx
x
4
+
d
dx

17x
12

+
d
dx
(37)
= 2
d
dx
x
3
+
d
dx
x
4
17
d
dx
x
12
+ 0
= 23x
2
+ 4x
3
1712x
11
= 6x
2
+ 4x
3
204x
11

27. Derivatives of polynomials
Example
Find
d
dx

2x
3
+x
4
17x
12
+ 37

Solution
d
dx

2x
3
+x
4
17x
12
+ 37

=
d
dx

2x
3

+
d
dx
x
4
+
d
dx

17x
12

+
d
dx
(37)
= 2
d
dx
x
3
+
d
dx
x
4
17
d
dx
x
12
+ 0
= 23x
2
+ 4x
3
1712x
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 ofx7!2
x
Example
Letf(x) = 2
x
. Use a calculator to estimatef
0
(0).
Solution
We have
f
0
(0) = lim
h!0
2
0+h
2
0
h
= lim
h!0
2
h
1
h
0:693147

30. Derivative ofx7!2
x
Example
Letf(x) = 2
x
. Use a calculator to estimatef
0
(0).
Solution
We have
f
0
(0) = lim
h!0
2
0+h
2
0
h
= lim
h!0
2
h
1
h
0:693147

31. Example
Use the previous fact to nd
d
dx
2
x
.
Solution
d
dx
2
x
= lim
h!0
2
x+h
2
x
h
= lim
h!0
2
x
2
h
2
x
h
= lim
h!0
2
x

2
h
1
h
= 2
x
lim
h!0
2
h
1
h
(0:693147)2
x
Here we have a function whose derivative is a multiple of itself!
(Much dierent from a polynomial.)

32. Example
Use the previous fact to nd
d
dx
2
x
.
Solution
d
dx
2
x
= lim
h!0
2
x+h
2
x
h
= lim
h!0
2
x
2
h
2
x
h
= lim
h!0
2
x

2
h
1
h
= 2
x
lim
h!0
2
h
1
h
(0:693147)2
x
Here we have a function whose derivative is a multiple of itself!
(Much dierent from a polynomial.)

33. Example
Use the previous fact to nd
d
dx
2
x
.
Solution
d
dx
2
x
= lim
h!0
2
x+h
2
x
h
= lim
h!0
2
x
2
h
2
x
h
= lim
h!0
2
x

2
h
1
h
= 2
x
lim
h!0
2
h
1
h
(0:693147)2
x
Here we have a function whose derivative is a multiple of itself!
(Much dierent from a polynomial.)

34. Example
Find
d
dx
3
x
.
Solution
d
dx
3
x
= lim
h!0
3
x+h
3
x
h
= lim
h!0
3
x
2
h
3
x
h
= lim
h!0
3
x

3
h
1
h
= 3
x
lim
h!0
3
h
1
h
(1:09861)3
x

35. Example
Find
d
dx
3
x
.
Solution
d
dx
3
x
= lim
h!0
3
x+h
3
x
h
= lim
h!0
3
x
2
h
3
x
h
= lim
h!0
3
x

3
h
1
h
= 3
x
lim
h!0
3
h
1
h
(1:09861)3
x

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

37. The natural exponential function
IIfa= 2,
d
dx
a
x




x=0
<1
IIfa= 3,
d
dx
a
x




x=0
>1
IWe would hope there is a numberabetween 2 and 3 such
that
d
dx
a
x




x=0
= 1
IWe call this numbere. Then by denition
d
dx
e
x
=e
x

38. Example
Find
d
dx

4x
2
+
1
x
+ 3
4
p
x+ 6e
x

Solution
Remember
1
x
=x
1
and
4
p
x=x
1=4
. So
dy
dx
= 8x
1
x
2
+
3
4
x
3=4
+ 6e
x

39. Example
Find
d
dx

4x
2
+
1
x
+ 3
4
p
x+ 6e
x

Solution
Remember
1
x
=x
1
and
4
p
x=x
1=4
. So
dy
dx
= 8x
1
x
2
+
3
4
x
3=4
+ 6e
x