Transcript 5.5 Differentiation of The Log Function

5.5 Differentiation of
Logarithmic Functions
By
Dr. 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
d
1
ln x 
dx
x
x  0 
Let’s see if we can discover why the rule is as above.
First define the natural log function as follows:
Now rewrite in exponential form:
Now differentiate implicitly:
ey  x
e y y  1
y 
1
1

y
x
e
y  lnx 
Example 1: Find the derivative of f(x)= xlnx.
Solution: This derivative will require the product rule.
f(x)  xlnx
1
f (x)  x  lnx1
x
f (x)  1  lnx
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.
lnx
x
1
x
 l n x 1
x
g (x) 
x2
g (x) 
g (x) 
1  lnx
x2
Quotient Rule:
(bottom)(derivative of top) – (top)(derivative of bottom)
(bottom)²
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.
Good work! Using the product rule:
F’(x) = (1st)(derivative of 2nd) + (2nd)(derivative of
1st)
1
y’ = x²  x  + (lnx)(2x)
 
y’ = x + 2xlnx
This can also be written y’ = x(1+2lnx)
Here is the second rule for differentiating logarithmic functions.
Rule 4: The Chain Rule for Log Functions
d
ln f(x)  f(x)
dx
f(x)
f(x)  0
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 f(x)  ln(x2  1)
Solution: Using the chain rule for logarithmic functions.
f(x)  ln(x2  1)
2x

f (x )  2
x 1
Derivative of the inside, x²+1
The inside, x²+1

Example 4: Differentiate y  ln x2  1x3  2
6

Solution: There are two ways to do this problem.
One is easy and the other is more difficult.
The difficult way:



6
d  2
3
5
6
x 1 x  2
2
3
2
3


x  1 6 x  2 3x  x  2 2x


y   dx

 x 2 1 x 3  2 6 
 x 2 1 x 3  2 6 





y 
y 






5


18x2 x 2  1 x 3  2  2x x 3  2

x
x
2

1 x 3  2


6

6




 
x  1x
5

 2
2x x 3  2 9x x 2  1  x 3  2
2
3
6
  2x10x  9x  2  20x  18x  4x
 1x  2
x  1 x  2 x  1x  2
2x 9x 3  9x  x 3  2
2



3
3
2
2
4
3
7
2
2
3



y  ln x 2  1 x 3  2

6
The easy way requires that we simplify the log using some of
the expansion properties.




 




6
6
y  ln x 2  1 x 3  2   ln x 2  1  ln x 3  2  ln x 2  1  6ln x 3  2


Now using the simplified version of y we find y ’.



y  ln x 2  1  6ln x 3  2
y 
 
x  1 x  2
2x
2

6 3x 2

3
Now get a common denominator.
y 

  
x  1x  2 x  2x  1
2x x 3  2
2
3


6 3x 2 x 2  1
3
2

Now that you have a common denominator, combine into a single
fraction.
y 
y 
y 

  
x  1x  2 x  2x  1
2x x 3  2
2
x

3
2x 4  4x
2

1 x  2
3


6 3x 2 x 2  1
3
2
18x4  18x2
 x
3


 2 x 2 1
20x4  18x2  4x
x
2

1 x 3  2

You’ll notice this is the same as the first solution.

Example 6: Differentiate gt   ln t e
2
 t2

Solution: Using what we learned in the previous example.
Expand first:

gt   ln t e
2
 t2
  ln t
2
 ln e
 t2
 2 ln t  t2
Now differentiate:
Recall lnex=x
gt   2 ln t  t2
2
g(t) 
 2t
t
 x4
Find the derivative of y  ln
 .
 x 3
Following the method of the previous two examples.
What is the next step?
Expanding to y  lnx  4  lnx  3
1
d  x4
Differentiating to y' 



x  4 dx  x  3 
x -3
This method of differentiating is valid, but it is the more difficult
way 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
y  lnx  4  lnx  3
Then find the derivative using the rule 4 on each logarithm.
1
1
y'

x  4 x 3
Now get a common denominator and simplify.
y' 
x -3
x4

x  4x  3 x  4x  3
y' 
7
x  4x  3
Example 7: Differentiate y  x(x  1)(x2  1)
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.
ln y  ln x(x  1)(x2  1)
Step 2 Expand the complicated side.
ln y  ln x(x  1)(x2  1)
ln y  ln x  ln(x  1)  ln(x2  1)
Step 3 Differentiate both side (implicitly for ln y )
ln y  ln x  ln(x  1)  ln(x2  1)
y 1
1
2x
 
 2
y x x 1 x 1
y 1
1
2x
 
 2
y
x x 1 x 1
Step 4: Solve for y ‘.
1
2x 
1

y  y 
 2

 x x 1 x 1 
Step 5:Substitute y in the above equation and simplify.
y  x(x  1)(x2  1)
1
2x 
1
y   x x  1(x 2  1) 
 2 
 x x 1 x 1


 x x  1(x 2  1) x x  1(x 2  1) 2x x x  1(x 2  1) 

y  




x
x 1
x 2 1


Continue to simplify…


 x x  1(x 2  1) x x  1(x 2  1) 2x x x  1(x 2  1)
y  


2

x
x

1
x
1



y   x  1(x 2  1)  x(x 2  1)  2xx x  1

y   x 3  x 2  x  1  x 3  x  2x 3  2x 2
y   4x 3  3x 2  2x  1



Let’s double check to make sure that derivative is correct by
Multiplying out the original and then taking the derivative.
y  x(x  1)(x2  1)


y  x2  x (x2  1)  x 4  x3  x2  x
y  4x3  3x2  2x  1
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.
ln y  ln x x
ln y  x ln x
Now, to find the derivative we differentiate both sides implicitly.
ln y  x ln x
y
1
 x  ln x  1
y
x
y
 1  ln x
y
y  y1  ln x   x x 1  ln x 
To find the critical numbers, set y’ = 0 and solve for x.
y  y1  ln x   x x 1  ln x 
y
0  x x 1  ln x 
0  1  ln x
 1  ln x
e 1  x
x
e 1 
1
 .367 ...
e
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
Thus, the minimum point occurs at x = 1/e or about .37
We learned two rules for differentiating logarithmic functions:
Rule 3: Derivative of ln x
d
1
ln x 
dx
x
x  0 
Rule 4: The Chain Rule for Log Functions
d
ln f(x)  f(x)
dx
f(x)
f(x)  0
We also learned it can be beneficial to expand a logarithm before
you take the derivative and that sometimes it is useful to take the
natural log (ln) of both sides of an equation, rewrite and then take
the derivative implicitly.