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Review 4.1-4.3 Differentiation of
Exponential Functions
Solving Exponential Equations
– Law 3 of the Laws of Logarithms says that: logaAC = C logaA
(Law 3)
2 7
ln2 ln7
ln2 ln7
ln72.807
ln2
x
x
x
x
Example 1: Find the derivative of f(x) = x2ex .
Solution: Do you remember the product rule? You will need it here.
2xeex(x)f
exf(x)xx2
x2
Product Rule:(1st)(derivative of 2nd) + (2nd)(derivative of 1st)
2xxe(x)f x Factor out the common factor xex.
Example 2: Find the derivative of f(t) = 23
t 2e
Solution: We will need the chain rule for this one.
t2
1t
23
t
e2e23
tf
2etf
)(
)(
Chain Rule:(derivative of the outside)(derivative of the inside)
The quotient rule results in . 4
xx2
x
2xeexx'f
Now simplify the derivative by factoring the numerator and canceling.
3
x
4
x
4
xx2
x
2xex'f
x
2xxe
x
2xeexx'f
Find the derivative of . 2
x
x
exf
Example 3: Find the derivative of f(x) = x3e
Solution: We will have to use Rule 2. The exponent, 3x is a function of x whose derivative is 3.
3exf
exfx3
x3
)(
)(
An exact copy ofthe exponential function
Times the derivative ofthe exponent
Example 4: Find the derivative of 1x2 2
exf )(
Solution:
4xe(x)f
ef(x)
12x
12x
2
2
12x2
4xe(x)f
Again, we used Rule 2. So the derivative is the exponential function times the derivative of the exponent.
Or rewritten:
Example 5: Differentiate the function tt
t
eee
tf )(
2tt
tttttt
ee
eeeeee(t)f
Solution: Using the quotient rule
2tt
02t02t
t
ee
eeee(t)f
s.'theintoeDistribute
Keep in mind that thederivative of e-t is e-t(-1) or -e-t
Recall that e0 = 1.
2tt ee
2(t)f
Here is the derivative in detail.
5x2
5ex'f
5x2
5ex'f
55x2
1ex'f
5xd
ex'f
5x
5x
2
1-5x
5x
dx
Find the derivative of . 5xexf
Example 6: A quantity growing according to the law where Q0 and k are positive constants and t
belongs to the interval experiences exponential growth.
Show that the rate of growth Q’(t) is directly proportional to the amount of the quantity present.
kt0eQtQ )(
,0
Solution:
)()(
)(
tkQkeQtQ
eQtQkt
0
kt0
Remember: To say Q’(t) is directly proportional to Q(t) means that for some constant k, Q’(t) = kQ(t) which was easy to show.
Example 7: Find the inflection points of 2xexf )(
Solution: We must use the 2nd derivative to find inflection points.
22
21
x
21
x
1x2
0e2
1x2e20
e2ex4xf
2ex2ex2xf
xe2xf
exf
2
2
x
2x
xx2
xx
x
x
2
2
22
22
2
2
)(
)(
)(
)(
First derivative
Product rule for second derivative
SimplifySet equal to 0.
Exponentials never equal 0.
Set the other factor = 0.
Solve by square root of both sides.
To show that they are inflection points we put them on a number line and do a test with the 2nd derivative:
72
2.
7
22
.
Intervals Test Points Value
,
,
,
22
22
22
22 -1
0
1
f”(-1)= 4e-1-2e-1=2e-1=+
f”(0)=0-2=-2 = -
f”(1)= 4e-1-2e-1=2e-1=+
22
2
xx2
x
e2ex4xf
exf
)(
)(
+ - +
Since there is a sign change across the potential inflection points,
21
e2
2, and are inflection points.
21
e22
,
In this lesson you learned two new rules of differentiation and used rules you have previously learned to find derivatives of exponential functions.
The two rules you learned are:
Rule 1: Derivative of the Exponential Function
xx eedx
d
Rule 2: If f(x) is a differentiable function then
)()()( xfeedxd xfxf
Find dy/dx
• x3 + y3 = 9xy
2 2dy dy3y 9x 9y 3x
dx dx
2 2 dy dy3x 3y 9x 9y
dx dx
2 2dy(3y 9x) 9y 3x
dx
2
2
dy 9y 3xdx 3y 9x
2
2
2
2
3(3y x )
3(y 3
3y x
3x) y x
