5.1 The Natural Logarithmic Function: Differentiation

5.1 The Natural Logarithmic Function: Differentiation

After this lesson, you should be able to:

Develop and use properties of the natural logarithmic function.Understand the definition of the number e.Find derivatives of functions involving the natural logarithmic function.

Now we will find derivatives of logarithmic functions and we will need rules for finding their derivatives.

Derivative of ln x 0x

Let’s see if we can discover why the rule is as above.

xy lnFirst define the natural log function as follows:

xe y

Now differentiate implicitly:

xey

ye

y

y

11

1

Now rewrite in exponential form:

xx

dx

d 1ln

Theorem 5.1 Properties of the Natural Logarithmic Function

Example 1 Find the derivative of f(x)= xlnx.

Solution This derivative will require the product rule.

1ln1

)('

ln)(

xxxxf

xxxf

xxf ln1)(

Product Rule:

(1st)(derivative of 2nd) + (2nd)(derivative of 1st)

Example 2 Find the derivative of g(x)= lnx/x

Solution This derivative will require the quotient rule.

2

1ln1

)(

ln)(

x

xxx

xg

x

xxg

Quotient Rule:

(bottom)(derivative of top) – (top)(derivative of bottom)

(bottom)²

2

ln1)(

x

xxg

Why don’t you try one: Find the derivative of y = x²lnx .

The derivative will require you to use the product rule.

Which of the following is the correct?

y ’ = 2

y ’ = 2xlnx

y ’ = x + 2xlnx

No, sorry that is not the correct answer.

Keep in mind - Product Rule:

(1st)(derivative of 2nd) + (2nd)(derivative of 1st)

Try again.

F ’(x) = (1st)(derivative of 2nd) + (2nd)(derivative of 1st)

Good work! Using the product rule:

y ’ = x² + (lnx)(2x)

y ’ = x + 2xlnx

This can also be written y’ = x(1+2lnx)

x

1

Theorem 5.2 Logarithmic Properties

The Chain Rule for Log Functions

)(

)()(ln

xf

xfxf

dx

d 0)( xf

Here is the second rule for differentiating logarithmic functions.

In words, the derivative of the natural log of f(x) is 1 over f(x) times the derivative of f(x)

Or, the derivative of the natural log of f(x) is the derivative of f(x) over f(x)

Theorem 5.3 Derivative of the Natural Logarithmic Function

Example 3 Find the derivative of )1ln()( 2 xxf

Solution Using the chain rule for logarithmic functions.

1

2)(

)1ln()(

2

2

x

xxf

xxf

Derivative of the inside, x²+1

The inside, x²+1

Example 4 Differentiate 632 )2)(1(ln xxy

Solution There are two ways to do this problem. One is easy and the other is more difficult. The difficult

way:

21

41820

21

29102

21

2992

21

21922

21

222118

21

223261

21

21

32

24

7322

3

32

33

632

3253

632

635322

632

632532

632

632

xx

xxx

xx

xxx

xx

xxxx

xx

xxxxx

xx

xxxxx

xx

xxxxx

xx

xxdxd

y

The easy way requires that we simplify the log using some of the expansion properties.

2ln61ln2ln1ln)2)(1(ln 32632632 xxxxxxy

Now using the simplified version of y we find y ’.

2

36

1

2

2ln61ln

3

2

2

32

x

x

x

xy

xxy

632 )2)(1(ln xxy

12

136

21

2223

22

32

3

xx

xx

xx

xxy

Now that you have a common denominator, combine into a singlefraction.

21

41820

12

1818

21

42

12

136

21

22

32

24

23

24

32

4

23

22

32

3

xx

xxx

xx

xx

xx

xx

xx

xx

xx

xxy

Now we can simplify

You’ll notice this is the same as the first solution.

Example 5 Differentiate )ln(22 tettg

222 ln2lnln)ln()(22

ttetettg tt

Solution Using what we learned in the previous example.Expand first:

Now differentiate:

tt

tg

tttg

22

)(

ln2)( 2

Recall lnex=x

Now you try: Find the derivative of .

3

4lnx

xy

Following the method of the previous two examples.What is the next step?

3xln4xlnytoExpanding

3x

4x

dx

d

3-x

4x1

y'toatingDifferenti

This method of differentiating is valid, but it is the more difficult way to find the derivative.

It would be simpler to expand first using properties of logs and then find the derivative.

Click and you will see the correct expansion followed by the derivative.

Correct. First you should expand to 3ln4ln xxy

Then find the derivative using the rule 4 on each logarithm.

3

1

4

1

xxy'

Now get a common denominator and simplify.

34

7

34

4

34

3

xxy'

xx

x

xx

xy'

Example 6 Differentiate )1)(1( 2 xxxy

Solution Although this problem could be easily done by multiplying the expression out, I would like to introduce to you a technique which you can use when the expression is a lot more complicated. Step 1 Take the natural log ln of both sides.)1)(1(lnln 2 xxxy

Step 2 Expand the complicated side.

)1ln()1ln(lnln

)1)(1(lnln2

2

xxxy

xxxy

Step 3 Differentiate both side (implicitly for ln y )

1

2

1

11

)1ln()1ln(lnln

2

2

x

x

xxy

y

xxxy

1

2

1

112

x

x

xxy

y

Step 4: Solve for y ’.

1

2

1

112x

x

xxyy

)1)(1( 2 xxxy

Step 5: Substitute y in the above equation and simplify.

1

)1)(1(2

1

)1)(1()1)(1(

1

2

1

11)1)(1(

2

222

22

x

xxxx

x

xxx

x

xxxy

x

x

xxxxxy

1234

221

)1(2)1()1)(1(

23

23323

22

xxx

xxxxxxx

xxxxxxx

Continue to simplify…

1

)1)(1(2

1

)1)(1()1)(1(2

222

x

xxxx

x

xxx

x

xxxy

Let’s double check to make sure that derivative is correct byMultiplying out the original and then taking the derivative.

1234

)1(

)1)(1(

23

23422

2

xxxy

xxxxxxxy

xxxy

Remember this problem was to practice the technique. You would not use it on something this simple.

Example 7 Consider the function y = xx.

Not a power function nor an exponential function.

This is the graph: domain x > 0

What is that minimum point?

Recall to find a minimum, we need to find the first derivative, find the critical numbers and use either the First Derivative Test or the Second Derivative Test to determine the extrema.

To find the derivative of y = xx , we will take the ln of both sides first and then expand.

xxy

xy x

lnln

lnln

Now, to find the derivative we differentiate both sides implicitly.

xxxyy

xy

y

xxx

y

y

xxy

x ln1ln1

ln1

1ln1

lnln

1

ln1

ln10

ln10

ln1ln1

ex

x

x

xx

xxxyyx

x

To find the critical numbers, set y ’ = 0 and solve for x.

...367.011

eex

Thus, the minimum point occurs at x = 1/e or about 0.367

x

y

Now test x = 0.1 in y ’, y ’(0.1) = -1.034 < 0 and x = 0.5 in y ’, y ’(0.5) = 0.216 > 0

Summary

We learned two rules for differentiating logarithmic functions:

Rule 1: Derivative of lnx

xx

dx

d 1ln 0x

Rule 2: The Chain Rule for Log Functions

)(

)()(ln

xf

xfxf

dx

d 0)( xf

We also learned it can be beneficial to expand a logarithm before you take the derivative and that sometimes it is useful to take thenatural log (ln) of both sides of an equation, rewrite and then takethe derivative implicitly.

Homework

Section 5.1 Pg. 329 7-33 odds, 45-59 odds, 71, 75, 77, 79, 83, 85