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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.