By
Dr. Julia Arnold and Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

Let’s consider the derivative of the exponential function.
Going back to our limit definition of the derivative: d dx
 f

 h lim

0 f
 x
 h
   h d dx
First rewrite the exponential using exponent rules.
Next, factor out e x.
Since e x does not contain h, we can move it outside the limit.
 h lim

0 e x
 h h
 e x
 h lim

0 e x e h h
 e x
 h lim

0 e x h h 
 e x h lim

0 e h h

1

Substituting h=0 in the limit expression results in the
0 indeterminate form , thus we will need to determine it.
f
( x
)
 x
We can look at the graph of and observe what happens as x gets close to 0. We can also create a table of values close to either side of 0 and see what number we are closing in on.
Graph
Table x -.1 -.01 -.001 .001 .01 .1
y .95 .995 .999 1.0005 1.005 1.05
At x = 0, f(0) appears to be 1.
As x approaches 0, y approaches 1.

We can safely say that from the last slide that lim h

0 e h h

1

1
Thus d dx e
 e x h lim

0 h

 e x 
1
 e x
Rule 1: Derivative of the Exponential Function d dx e
 
 e x
The derivative of the exponential function is the exponential function.

Find the derivative of f(x) = x 2 e x .
Solution: Do you remember the product rule? You will need it here.
Product Rule:
(1 st )(derivative of 2 nd ) + (2 nd )(derivative of 1 st ) f(x)
 x
2 e x f

(x)
 x
2 e x  e x
2x
Factor out the common factor xe x .
f

(x)
 xe x
 x

2


Find the derivative of f(t) =
 e t  2
 3
2
Solution: We will need the chain rule for this one.
Chain Rule:
(derivative of the outside)(derivative of the inside) f
( t
) f 
( t
)


 e t  2
 3
2
3
2
 e t  2
1 
2 e t

Why don’t you try one: Find the derivative of f
 
 e x x
2
.
To find the solution you should use the quotient rule.
Choose from the expressions below which is the correct use of the quotient rule.
f ' f '
 e x
2x
 x
2 e x  e x x
4 f '
 e x  x
2 e x x
4

No that’s not the right choice.
Remember the Quotient Rule:
(bottom)(derivative of top) – (top)(derivative of bottom)
(bottom)²
Try again.
Return

Good work!
The quotient rule results in f '
 x
2 e x  e x x
4
.
Now simplify the derivative by factoring the numerator and canceling.
f ' f '
 x
2 e x 
2xe x x
4
 xe x
 x

2
 x
4
 e x
 x

2
 x
3

What if the exponent on e is a function of x and not just x?
Rule 2: If f(x) is a differentiable function then d dx
 e f
( x
)  f 
( x
)
In words: the derivative of e to the f(x) is an exact copy of e to the f(x) times the derivative of f(x).

Find the derivative of f(x) = e 3 x
Solution: We will have to use Rule 2.
The exponent, 3x is a function of x whose derivative is 3.
f(x)
 e
3x f

(x)
 e
3x 
3

3e
3x
An exact copy of the exponential function
Times the derivative of the exponent

Find the derivative of f
( x
)
 e 2 x 2  1
Solution: f(x)
 e
2x
2 
1 f

(x)
 e
2x
2 
1 Again, we used Rule 2. So the derivative is the exponential function times the derivative of the exponent.
Or rewritten: f

(x)

4xe
2x
2 
1

Differentiate the function f
( t
)
 e t e t
 e  t
Solution: Using the quotient rule f

(t)
 e
 t  e
 t e
 t e
 t

 e e
 t t
2 e

 t  e
 t

Keep in mind that the derivative of e -t is e -t (-1) or -e -t
Distribute e t into the
 
' s.
f

(t)
 e
2t  0

 2t e
 e t e e
 t

2
 e
0 f

(t)
 e
 t 
2 e
 t

2
Recall that e 0 = 1.

Find the derivative of f
 
 e
5x .
Click on the button for the correct answer.
f '

5e
2
5x
5x f '
 e
5 x
5 x

No, the other answer was correct.
Remember when you are doing the derivative of e raised to the power f(x) the solution is e raised to the same power times the derivative of the exponent.
What is the derivative of
5x
?
Try again.
Return

Good work!!
Here is the derivative in detail.
f ' f ' f ' f '


 e e e
5x  d dx
5x 
1
2
5
5x 
2 5x

5e
2
5x
5x

A quantity growing according to the law
Q
( t
)
 Q
0 e kt where Q
0 and k are positive constants and t
 
Show that the rate of growth Q’(t) is directly proportional to the amount of the quantity present.
Solution: Q
( t
)
Q 
( t
)


Q
0 e kt
Q
0 e kt  k  kQ
( t
)
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.

Find the inflection points of f
( x
)
 e  x 2
Solution: We must use the 2 nd derivative to find inflection points. f
( x
)
 e  x 2 f 
( x
) f 
( x
)


 2 xe  x 2

First derivative

 2 x
 e  x 2 
 2 x


 e  x 2   f
0

( x
)
 2 e

 x 2
4 x 2
 e
2 x

2 x 2
 1


2 e  x 2
Simplify
Set equal to 0.
Product rule for second derivative
2 e  x 2 
2 x 2 x 2 
 1
1
2 x  
1
2
0 Exponentials never equal 0.
Set the other factor = 0.
 
2
2
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 2 nd derivative: f
( x
)
 e  x 2 f 
( x
)
 4 x 2 e  x 2  2 e  x 2
+ +

2
2
 
.
7
2
2

.
7
Intervals Test Points Value





 
,

2
2



 
2
2
,
2
2



-1
0 f”(-1)= 4e -1 -2e -1 f”(0)=0-2=-2 = -
=2e -1 =+



2
2
,



 1 f”(1)= 4e -1 -2e -1 =2e -1 =+
Since there is a sign change across the potential inflection points,
 
 2
2
, e

1
2



 2
2
, e

1
2



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 d dx e
 
 e x
Rule 2: If f(x) is a differentiable function then d dx
 e f
( x
)  f 
( x
)