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Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Do Now:
Aim: How do we differentiate and integrate the exponential function?
1'( ) , '( ( )) 0
'( ( ))g x f g x
f g x
1 11
1'( ) , '( ( )) 0
'( ( ))f x f f x
f f x
1 3Find ' 2 for ( ) 2 1f f x x x
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Do Now
1 3Find 2 for ( ) 2 1f f x x x
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
The Natural Exponential Function
4
2
-2
-4
-5 5
f x = ln x
Characteristics of Natural Log Function
Monotonic - increasingDomain – (0, )Range – all reals
Has an inverse f -1
f -1(x) = ex
Natural Exponential
Function
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
4
2
-2
-4
-5 5
f x = ln x
Definition of Natural Exponential Function
Natural Log Function
f -1(x) = ex
Natural Exponential
Function 1 1lnf f x f x x
The inverse of the natural logarithmic function f(x) = ln x is called the natural
exponential function and is denoted by f -1(x) = ex. That is, y = ex x = ln y
ln(ex) = x ln e = x(1) = x
if x is rational
ln ex = x
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
4
2
-2
-4
-5 5
f x = ln x
Properties of Natural Exponential Function
Natural Log Function
f -1(x) = ex
Natural Exponential
Function
1. domain – (-, ); range – (0, )
2. continuous, increasing, and 1-to-1
3. concave up on its entire domain
4. lim 0 and limx x
x xe e
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Problems
Solve 4e2x = 5 to 3 decimal places
x 1
2ln
5
4
x 1
2.2231435513 0.112
ln e2x = ln 5/4 Property of Equality for Ln functions
2x = ln 5/4Inverse Property of Logs & Expos
e2x = 5/4 Divide both sides by 4
Check: 4e2(0.112) = 5
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Solving Exponential Equations
1Solve 7 xe
1ln7 ln xe take ln of both sides
ln7 1x
ln7 1 lnx e
ln7 1 x
0.946x
solve for x
apply inverse propertyln xe x
ln e = 1
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Solving Log Equations
Solve ln 2 3 5x
ln 2 3 5xe e expo both sides
52 3x e
513
2x e
75.707x solve for x
apply inverse propertyln xe x
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Complicated Problem
Solve e2x – 3ex + 2 = 0
Quadratic Form(ex)2 – 3ex + 2 = 0
Factor(ex – 2)(ex – 1) = 0
Set factors equal to zero(ex – 2) = 0 (ex – 1) = 0
ex = 2 ex = 1
x = ln 2 x = 0
x = 0.693 x = 0
Graph to verify
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Derivatives of Exponential Functions
Let be a differentiable function of
1. 2.x x u u
u x
d d due e e e
dx dx dx
2 1. xda e
dx
u due
dx
2 1 2 12 2x xe e
u = 2x - 1u’ = 2
3/. xdb e
dx
u due
dx
3/3/
2 2
3 3 xx e
ex x
u = -3/xu’ = 3/x2
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Model Problem
Find the relative extrema of f(x) = xex
'( ) 1x xf x x e e
1 1 0x x xx e e e x
4
2
-5
g x = ex
ex is never 0 x + 1 = 0
x = -1
4
2
-2
-5
h x = xex
A
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Model Problem
2 /2
Show that the normal probability density function
1( ) has points of inflection when 1.
2xf x e x
When 2nd derivative equals zero.
2 /21'( )
2xf x x e
u ud due e
dx dx
u = -x2/2; u’ = -x
2 2/2 /21''( ) 1
2x xf x x x e e
2 22 /2 /211 0
2x xx e e
2 /2 211 0
2xe x
x = ±1
0.5
0.4
0.3
0.2
0.1
-0.1
-2 2
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Model Problem
For 1980 through 1993, the number y of medical doctors in the U.S. can be modeled by y = 476,260e0.026663t where t = 0 represents 1980. At what rate was the number of M.D.’s changing in 1988?
When t in 1st derivative equals 8.
0.026663'( ) 0.026663 476,260 tf x e
0.026663 80.026663 476,260 e
15,718 doctors per year
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Integrals of Exponential FunctionsLet be a differentiable function of .
1. 2. x x u u
u x
e dx e C e du e C 3 1Evaluate xe dx
3 1 3 113
3x xe dx e dx
u = 3x + 1du/dx = 3;du = 3dx
multiple and divide by 3
1
3ue du
1
3ue C u ue du e C 3 11
3xe C Back-substitute
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Model Problem2
Evaluate 5 xxe dxu = -x2
du/dx = -2x du = -2xdx xdx = du/2
u ue du e C
2 2
5 5x xxe dx e xdx regroup integrand
52
u due
substitute
5
2ue du factor out -5/2
u ue du e C 5
2
ue C
25
2xe C Back-substitute
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Model Problem1
2
xedx
x u = 1/x 2 2
1 1dudu dx
dx x x
1
2
xedx
x
12
1u
u
x
de
e dxx
1/ xe C
cossin xx e dx u = cosx sin sindu
x du xdxdx
cos sin
u
x
due
e xdx
cos xe C
cossin xx e dx
Aim: Differentiating & Integrating Expo Functions
Course: Calculus
Model Problem
Find the areas1
0
xe dx 1
01 xe
1 1e
11 0.632
e
1.2
1
0.8
0.6
0.4
0.2
-0.2
0.5 1
t x = e-x
1
0 1
x
x
edx
e 1
0ln 1 xe
ln 1 ln2e
0.620
1.2
1
0.8
0.6
0.4
0.2
-0.2
0.5 1
u x = ex
1+ex
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