Differentiation of Logarithmic Functions

7/31/2019 Differentiation of Logarithmic Functions

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5.5 Differentiation of

Logarithmic FunctionsBy

Dr. Julia Arnold and Ms. Karen Overman

using Tans 5th edition Applied Calculus for themanagerial , life, and social sciences text


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

Rule 3: Derivative of ln x

0x

Lets see if we can discover why the rule is as above.

xy lnFirst define the natural log function as follows:

xey

Now differentiate implicitly:

x

1

e

1

y

1ye

y

y

Now rewrite in exponential form:

x

1x

dx

dln


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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)


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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)

2x

lnx1(x)g


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Why dont you try one: Find the derivative of y = xlnx .

The derivative will require you to use the product rule.

Which of the following is the correct?

y = 2

y = 2xlnx

y = x + 2xlnx


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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.


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F(x) = (1st)(derivative of 2nd) + (2nd)(derivative of1st)

Good work! Using the product rule:

y = x + (lnx)(2x)

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

x

1


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Rule 4: The Chain Rule for Log Functions

)(

)()(ln

xf

xfxf

dx

d 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 off(x) over f(x)


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

Solution: Using the chain rule for logarithmic functions.

1xx2xf

1xxf

2

2

)(

)ln()(

Derivative of the inside, x+1

The inside,x+1


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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

2x1x2x1x9x2x2x

2x1x2x2x2x1x18xy

2x1x

2x2x3x2x61x

2x1x

2x1xdx

d

y

32

24

7322

3

32

33

632

3253

632

635322

632

632

532

632

632


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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


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2x1x4x18x20x

y

1x2x18x18x

2x1x4x2xy

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.

Youll notice this is the same as the first solution.


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Example 6: Differentiate 2t2ettg ln

2t2t2 tt2etettg 22 lnlnlnln

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

Now differentiate:

t2t

2

tg

tt2tg 2

)(

ln

Recall lnex=x


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Find the derivative of .

3

4ln

x

xy

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

3xln4xlnytoExpanding

3x

4xdxd

3-x

4x1y'toatingDifferenti


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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 andthen find the derivative.

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


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Correct.First you should expand to 3xln4xlny

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

3x

1

4x

1

y'

Now get a common denominator and simplify.

3x4x

7y'

3x4x4x

3x4x3-xy'


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Example 7: Differentiate ))(( 1x1xxy 2

Solution: Although this problem could be easily done bymultiplying the expression out, I would like to introduce to youa technique which you can use when the expression is a lot morecomplicated.Step 1 Take the ln of both sides.

))((lnln1x1xxy

2

Step 2 Expand the complicated side.

)ln()ln(lnln

))((lnln

1x1xxy

1x1xxy2

2

Step 3 Differentiate both side (implicitly for ln y )

1x

x2

1x

1

x

1

y

y

1x1xxy

2

2

)ln()ln(lnln


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1x

2x

1x

1

x

1

y

y2

Step 4: Solve for y .

1x

x2

1x

1

x

1yy

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

1

1)(x1xxy

2

222

2

2


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12x3x4xy

2x2xxx1xxxy

1xx2x1)x(x1)(x1xy

1x

1)(x1xx2x

1x

1)(x1xx

x

1)(x1xx

23

23323

22

2

222

y

Continue to simplify


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Lets 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. Youwould not use it on something this simple.


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Consider the function y = xx.

Not a power function nor anexponential 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 DerivativeTest or the Second Derivative Test to determine the extrema.


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

xxy

xy x

lnln

lnln

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

x1xx1yy

x1y

y

1xx

1x

y

y

xxy

x lnln

ln

ln

lnln


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xe

x1

x10

x1x0x1xx1yy

1

x

x

ln

ln

ln

lnln

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

....367e

1e 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 < 0and x = 0.5 in y, y(0.5) = 0.216 > 0


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We learned two rules for differentiating logarithmic functions:

Rule 3: Derivative of ln x

x

1x

dx

dln 0x

Rule 4: The Chain Rule for Log Functions

)(

)()(ln

xf

xfxf

dx

d 0xf )(

We also learned it can be beneficial to expand a logarithm beforeyou 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.

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Differentiation of Logarithmic Functions