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CHAPTER 5SECTION 5.4
EXPONENTIAL FUNCTIONS:DIFFERENTIATION AND
INTEGRATION
Definition of the Natural Exponential Function
Recall:
ln log where 2.718ex x e
ln and are inverse functionsxx e
xln e lnx e 1x x lnxe xThis means… and…
Exponential and log functions are interchangeable.
xy a logax y
Start with the base.
Change of Base Theorem
loga x log lnx or
log lna
x
a
10log logx x
Solve.16. 7 xe 7. ln 2 3 5x
3 38. log log 2 1x x
Solve.16. 7 xe 7. ln 2 3 5x
1ln7 ln xe ln7 1 lnx e
ln7 1x ln7 1 x 0.946 x
ln 2 3 5xe e 52 3x e 5 3
2
ex
75.707x
3 38. log log 2 1x x
3log 2 1x x 13 2x x 20 2 3x x
0 3 1x x 3, 1x 3x
We can’t take a log of -1.
Theorem 5.10 Operations with Exponential Functions
Properties of the Natural Exponential Function
Theorem 5.11 Derivative of the Natural Exponential Function
5.4 Exponential Functions
• Example 3: Find dy/dx:
a) y 3ex; b) y x2ex; c) y ex
x3 .
2 2
a) 3 3
3
b) 2
2
x x
x
x x x
x
dy de e
dx dxe
dx e x e e x
dxxe x
5.4 Exponential Functions
• Example 3 (concluded):
c) d
dx
ex
x3
x3 ex ex 3x2
x3 2
x2ex x 3
x6
ex (x 3)
x4
Find each derivative:
4
4
5
5
1( ) 2
2
' = 4 2 2
4 2
2
x
x
x x
x
x
f x x ex e
f x x e e
e
x e
5.4 Exponential Functions• THEOREM 2
•
• or
• The derivative of e to some power is the product of e • to that power and the derivative of the power.
d
dxe f (x ) e f (x ) f (x)
d
dxeu eu
du
dx
5.4 Exponential Functions• Example 4: Differentiate each of the
following with
• respect to x:a) y e8 x ; b) y e x2 4 x 7; c) y e x2 3 .
a) d
dxe8 x e8 x 8
8e8 x
b) d
dxe x2 4 x 7 e x2 4 x 7 2x 4
5.4 Exponential Functions
• Example 4 (concluded):
c) d
dxe x2 3
d
dxex2 3
1
2
ex2 3
1
2
1
2x2 3
1
2 2x
xe x2 3
x2 3
Find each derivative
74
64 4
4 4
2
7 2 4
28 2
x
x x
x x
y e
dye e
dx
e e
ude
dx Theorem:
122. x
de
dx
3. lnxde x
dx
1. Find the slope of the line tangent to f (x) at x = 3.
2xf x e
ude
dx Theorem:
122. x
de
dx
u due
dx
3. lnxde x
dx
1. Find the slope of the line tangent to f (x) at x = 3.
2' 2xf x e 2xf x e 6' 3 2f e
6
2
e
2 32xe x
2xde
dx
123
2xx e
1lnx xe e x
x
1
lnxe xx
1 1ln
xx
e x
1 lnx
x x
xe
4. Find extrema and inflection points for 2 xy x e
4. Find extrema and inflection points for 2 xy x e
2 2x xdyx e e x
dx
20 2 0xe x x Crit #’s:
Can’t ever work.
0,2x
2 2xe x x
2
22
2 2 2x xd ye x e x x
dx
22 2 2xe x x x
2 4 2xe x x Crit #’s:
20 4 2 0xe x x none
24 4 4 1 2
2 1x
4 8
2x
4 2 2
2x
2 2x
2 1.4x
.6,3.4x
Intervals: ,0 0,.6 .6,2
Test values: 1 12 1
f ’(test pt) f(x) dec inc inc
rel min rel max
0x 2x 0,0 2
42,e
f ’’(test pt) f(x) up up down
2,3.4 3.4,3 5
dec dec
down up
2 2xdye x x
dx
22
24 2xd y
e x xdx
2xe x x
Inf pt Inf pt
.6x 3.4x 0.586,0.191 3.414,0.384
2 xy x e
5.4 Exponential Functions
•Example 7: Graph with x ≥ 0. Analyze the graph using calculus.
•First, we find some values, plot the points, and sketch
•the graph.
h(x) 1 e 2 x
• Example 4 (continued):• a) Derivatives. Since
•
• b) Critical values. Since the derivative
for all real numbers x. Thus, the
• derivative exists for all real numbers, and the equation
• h(x) = 0 has no solution. There are no critical values.
h(x) 1 e 2 x ,
h (x) 2e 2 x
and
h (x) 4e 2 x .
h (x) 2e 2 x 0
e 2 x 1
e2 x 0,
•Example 4 (continued):•c) Increasing. Since the derivative for all real numbers x, we know that h is increasing over the entire real number line.
•d) Inflection Points. Since we know that the equation h(x) = 0 has no solution. Thus there are no points of inflection.
h (x) 2e 2 x 0
2( ) 4 0xh x e
5.4 Exponential Functions
• Example 4 (concluded):• e) Concavity. Since
for all real numbers x, h’ is decreasing and the graph is concave down over the entire real number line.
2( ) 4 0xh x e
• Example 4 (continued):
Theorem 5.12 Integration Rules for Exponential Functions
ue du Theorem:
2
1. xxe dx
2. 1x xe e dx
3.x x
x x
e edx
e e
ue du Theorem:
2
1. xxe dx
ue C
2. 1x xe e dx
2u x2du xdx
1 xu e xdu e dx
1
2ue du
1
2ue C
21
2xe C
u dx 3
22
3u C
322
13
xe C
3.x x
x x
e edx
e e
x xu e e
x xdu e e dx
1du
u ln x xe e C
1ln x
xe C
e
2 1ln
x
x
eC
e
2ln 1 lnx xe e C 2ln 1xe x C
u u u ud due e e du e C
dx dx
1
x
0
e 1 dx x 10e x | e 1 0 1 e
ln22x
0
e dxu 2x
1du dx
2
2ln2u
0
1e du
2
u 2ln2
0
1e |
2
11 1 3
2 2 84
u u u ud due e e du e C
dx dx
31
2 x x
0
3x 1 e dx
3
2
u x x
du 3x 1 dx
2u
0
e du2u
0e | 2e 1
ln22x
0
e dxu 2x
1du dx
2
2ln2u
0
1e du
2
u 2ln2
0
1e |
2
11 1 3
2 2 84
x u
x u u ud a d a du 1
a lna a lna a du a Cdx dx dx
Note : a 0lna
12x
0
3 dxu 2x
1du dx
2
2u
0
1du
23 u 2
0
13 |
2ln3 1
9 12ln3
8
ln9
0x
1
4 ln2dx
u x
du dx
0u
1
ln2 d4 u u 01
ln24 |
ln4
1
1 42
3
2
x u
x u u ud a d a du 1
a lna a lna a du a Cdx dx dx
Note : a 0lna
22
x
1
x2 dx2u x
1du xdx
2
2u
1
1du
22 u 2
1
12 |
2ln2 1
ln2
/ 3sec x
0
sec x tanx2 dx
u sec x
du sec x tanxdx
2u
1
2 du u 21
12 |
ln2 2
ln2
AP QUESTION
Why is x = -1/2 the only critical number???????
AP QUESTION
