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14

DERIVATIVES OF LOGARITHMIC

AND

EXPONENTIAL FUNCTIONS

The derivative of ln x

THE SYSTEM OF NATURAL LOGARITHMS has the number called e as it base; it is the system we use in all theoretical work. (In the next Lesson, we will see that e is approximately 2.718.) The system of natural logarithms is in contrast to the system of common logarithms, which has 10 as its base and is used for most practical work.

We denote the logarithmic function with base e as "ln x."

ln x  =  logex.

y = ln x   implies   e y = x.

In other words, this logarithm function --

y = ln x

-- has for its inverse the exponential function,

y = ex.

Here are the inverse relations:

ln ex = x   and   eln x = x.

And the logarithm of the base itself is always 1:

ln e = 1.

(Topic 20 of Precalculus.)

The function  y = ln x  is continuous and defined for all positive values of x.  It will obey the usual laws of logarithms:

1.    ln ab  =  ln a  +  ln b.

2.    ln  a
b
  =  ln a  −  ln b.

3.    ln an  =  n ln a.

(Topic 20 of Precalculus.)

Like all the rules of algebra, they will obey the rule of symmetry.
For example,

n ln a  =  ln an.

The derivative of ln x

We will now apply the definition of the derivative to prove:

 d 
dx
  ln x  =   1
x

In the course of the proof, it will make for a great simplification if we define the base of the system of natural logs, the number we call e, as the following limit:

e defined as a limit

A limit in the proof will have that same form.

Later, we will call the variable x rather than v. And in the next lesson, upon changing the variable from v to fraction, the familiar definition follows.

Here is the difference quotient:

The difference quotient   according to the 2nd law;
 =   Proof continued   on multiplying by x/x;
 
 =   Proof continued   according to the 3rd law.

We now take the limit as h approaches 0.

The derivative of ln x  =  The derivative of ln x
The limit does not apply to  1
x
, because h is the variable
   that is approaching 0.   1
x
 is a constant factor.

We now define that limit to be the base of the natural logarithms, the number we will call e.  (That limit is the one above, with v = fraction; when Right arrow0, fractionRight arrow0.)

Therefore,

The derivative of ln x  =   The derivative of ln x
 
   =   The derivative of ln x
 
   =   The derivative of ln x

Which is what we wanted to prove.

To see that this limit --

The limit that is e

-- that is, e, exists as x approaches 0, here is the graph of

Equation of the graph

The graph

y has a definite value as x approaches 0. And in the next Lesson we will see that it is approximately 2.718.

The derivative of ex

We will now prove:

 d 
dx
  ex  =  ex

"The derivative of ex with respect to x

is equal to ex."

Since y = ex is the inverse of y = ln x, we can obtain its derivative as follows:

  y  =  ex  
 
implies ln y  =  ln ex  = x.

Therefore on taking the derivative of both sides with respect to x, and applying the chain rule to ln y:

  The derivative of e to the x  =   1.
 
y'  =   y.
That is,
  The derivative of e to the x  =   ex.

ex is its own derivative.

What does that imply?  It implies the meaning of exponential growth.  For we say that a quantity grows "exponentially" when it grows at a rate that is proportional to its size. The bigger it is at any given time, the faster it's growing at that time. A typical example is population. The more individuals there are, the more births there will be, and hence the greater the rate of change of the population -- the number of births in each year.

All exponential functions have the form ax, where a is the base. Therefore, to say that the rate of growth is proportional to its size, is to say that the derivative of ax is proportional to ax.

 d 
dx
  ax  =  kax

where k is the constant of proportionality. (Lesson 39 of Algebra.) When we calculate that derivative below, we will see that that constant becomes ln a.

 d 
dx
  ax  =  ln a· ax .

In the system of natural logarithms, in which e is the base, we have the simplest constant possible, namely 1.

 d 
dx
  ex  =  ex.  

The derivative of e with a functional exponent

When  y = eu(x), then according to the chain rule:

The chain rule

That is,

The derivative

"The derivative of e with a functional exponent

is equal to e with that exponent
times the derivative of that exponent."

Example 1.   Calculate the derivative of e2x + 3.

  Solution.   2 times e to the 2x + 3.

Problem 1.   Calculate the derivative of ex2.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

ex2· 2x = 2x ex2

Problem 2.   Calculate the derivative of the following.

a)   esin x.    esin x cos x

b)  e−x.    ex (−1) = −ex

c)  x2ex.   x2ex + 2x ex

According to the product rule.

  d)   e to the x minus 1 over e to the x plus 1   According to the quotient rule:
The quotient rule = The quotient rule = The quotient rule

The derivative of ln u(x)

When  y = ln u(x), then according to the chain rule:

The chain rule

That is,

The chain rule

   Example 2.    Derivative of ln(2x + 3)
   Example 3.     d 
dx
  ln sin x  =      1   
sin x
· cos x  =   cos x
sin x
 =  cot x.

Example 4.   Find the derivative of ln x2.

Solution.   We may apply the laws of logarithms:

 d 
dx
  ln x2  =    d 
dx
2 ln x,   3rd law,
 
   =   2  d 
dx
ln x
 
   =   2
x
.
  Example 5.   Find the derivative of  ln      x    
3x − 4
.

Solution.   According to the 2nd Law:

 d 
dx
 ln      x    
3x − 4
  =    d 
dx
[ln x − ln (3x − 4)]
 
    =   Derivative of ln x squared
 
    =   Derivative of ln x squared
 
    =   2 over x

Problem 3.   Differentiate the following.

  a)   ln x3.     d 
dx
 ln x3  d 
dx
 3 ln x   =   3
x
  b)   (ln x)3.    3(ln x)2·  1
x
  =   3(ln x)2
    x
  c)   ln (3x2 − 4x).          1      
3x2 − 4x
· (6x − 4)   =     6x − 4  
3x2 − 4x
  d)   ln (3x − 4)2.          1      
(3x − 4)2
· 2(3x − 4)· 3   =   6(3x − 4)
(3x − 4)2
  =       6    
3x − 4
  e)   ln cos x.      1   
cos x
(−sin x)   =   sin x
cos x 
  =   −tan x
   Problem 4.   Calculate the derivative of ln  2
x
.
 d 
dx
ln  2
x
 =   d 
dx
(ln 2 − ln x) = 0 −  1
x
  =  − 1
x

Problem 5.  The derivative of logax.

According to the rule for changing from base e to a different base a:

Change of base

Topic 20 of Precalculus.

Derivative of log with base a

1 over x ln a

Calculate the limit of that derivative

a)  when x is greater than 1 and becomes larger.

That derivative approaches 0, that is, becomes smaller.

b)  when x is less than 1 and becomes smaller.

That derivative becomes larger.

The general power rule

We can now prove that the derivative of  f(x) = xn, where n is any rational exponent, is as follows:

 d 
dx
  xn  =  nxn−1

Let

  y  =  xn.
  Then
  ln y  =  n ln x  (3rd Law).
 
  Therefore, on taking the derivative with respect to x:
 
  Proof of the power rule  =  n
x
  so that
  y'  =  n
x
· y
 
     =  n
x
· xn
 
      =  nxn−1.

That is what we wanted to prove.

(If n is 0, then x0 = 1, a constant; its derivative is 0.  If n is irrational, a rational approximation will be necessary.)

   Problem 6.   Calculate the derivative of  Fifth root of ln 5x

Solution

The derivative of ax

We will prove:

 d 
dx
  ax  =  ln a· ax

"The derivative of an exponential function with base a

is equal to the natural logarithm of that base

times the exponential function."

Let

y = ax.
 
  Then on taking the natural logarithm of both sides:
 
ln y = x ln a.  (3rd Law)
 
  Therefore,
Derivative of a to the x = Derivative of a to the x
 
  But by the chain rule:
 
Derivative of a to the x = Derivative of a to the x
 
  Therefore,
 
Derivative of a to the x = ln a.
 
Derivative of a to the x = ln a
 
y' = ln a· y
 
  That is,
Derivative of a to the x = ln a· ax.

This is what we wanted to prove.

   Example 6.      d 
dx
  2x  =  ln 2· 2x.

Problem 7.   Calculate the derivative of  y = 105x.

By the chain rule:

Solution

End of the lesson

Next Lesson:  Evaluating e

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