5.5 Differentiation of Logarithmic Functions

ByDr. Julia Arnold and Ms. Karen Overman

using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

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

Rule 3: 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:

xey

Now differentiate implicitly:

x1

e1

y

1ye

y

y

Now rewrite in exponential form:

x1

xdxd

ln

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

Solution: This derivative will require the product rule.

1lnxx

1x(x)f

xlnxf(x)

lnx1(x)f

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.

2x

1lnxx

1x

(x)g

x

lnxg(x)

Quotient Rule:

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

(bottom)²

2x

lnx1(x)g

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. Return to previous slide.

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

Rule 4: The Chain Rule for Log Functions

)(

)()(ln

xfxf

xfdxd

0xf )(

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)

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

Solution: Using the chain rule for logarithmic functions.

1xx2

xf

1xxf

2

2

)(

)ln()(

Derivative of the inside, x²+1

The inside, x²+1

Example 4: Differentiate 632 2x1xy ln

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

2x1x

4x18x20x

2x1x

29x10x2x

2x1x

2x9x9x2xy

2x1x

2x1x9x2x2x

2x1x

2x2x2x1x18xy

2x1x

2x2x3x2x61x

2x1x

2x1xdx

d

y

32

24

7322

3

32

33

632

3253

632

635322

632

632532

632

632

632 2x1xy ln

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

2x6ln1xln2xln1xln2x1xlny 32632632

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

1x2x

1x3x6

2x1x

2x2xy

r.denominatocommonagetNow

2x

3x6

1x

2xy

2x6ln1xlny

23

22

32

3

3

2

2

32

2x1x

4x18x20xy

1x2x

18x18x

2x1x

4x2xy

1x2x

1x3x6

2x1x

2x2xy

32

24

23

24

32

4

23

22

32

3

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

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

Example 6: Differentiate 2t2ettg ln

2t2t2 tt2etettg22

lnlnlnln

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

Now differentiate:

t2t2

tg

tt2tg 2

)(

ln

Recall lnex=x

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 difficultway to find the derivative.

It would be simplier 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 3xln4xlny

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

3x

1

4x

1y'

Now get a common denominator and simplify.

3x4x

7y'

3x4x

4x

3x4x

3-xy'

Example 7: Differentiate ))(( 1x1xxy 2

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 ln of both sides.

))((lnln 1x1xxy 2

Step 2 Expand the complicated side.

)ln()ln(lnln

))((lnln

1x1xxy

1x1xxy2

2

Step 3 Differentiate both side (implicitly for ln y )

1xx2

1x1

x1

yy

1x1xxy

2

2

)ln()ln(lnln

1x

2x

1x

1

x

1

y

y2

Step 4: Solve for y ‘.

1xx2

1x1

x1

yy 2

))(( 1x1xxy 2

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

1x

1)(x1xx2x

1x

1)(x1xx

x

1)(x1xxy

1x

2x

1x

1

x

11)(x1xxy

2

222

22

12x3x4xy

2x2xxx1xxxy

1xx2x1)x(x1)(x1xy

1x

1)(x1xx2x

1x

1)(x1xx

x

1)(x1xx

23

23323

22

2

222

y

Continue to simplify…

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

1x2x3x4y

xxxx1xxxy

1x1xxy

23

23422

2

)(

))((

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

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.

x1xx1yy

x1yy

1xx1

xyy

xxy

x lnln

ln

ln

lnln

xe

x1

x10

x1x0

x1xx1yy

1

x

x

ln

ln

ln

lnln

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

....367e1

e 1

Thus, the minimum point occurs at x = 1/e or about .37

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

We learned two rules for differentiating logarithmic functions:

Rule 3: Derivative of ln x

x1

xdxd

ln 0x

Rule 4: The Chain Rule for Log Functions

)(

)()(ln

xfxf

xfdxd

0xf )(

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.