Logarithmic Differentiation

  • #1
Something about this worked problem looks off. Is this example correctly solved using logarithmic differentiation?

The original problem is y = (2-x)^(sqrt x). If anyone who is rather confident with this could double check this example it would really help me out. Thanks. I attached the link to reduce my typing confusion.

http://archives.math.utk.edu/visual.calculus/3/logdiff.1/a1.gif
 

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  • #2
SammyS
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Something about this worked problem looks off. Is this example correctly solved using logarithmic differentiation?

The original problem is y = (2-x)^(sqrt x). If anyone who is rather confident with this could double check this example it would really help me out. Thanks. I attached the link to reduce my typing confusion.

http://archives.math.utk.edu/visual.calculus/3/logdiff.1/a1.gif
Here's the image:
attachment.php?attachmentid=49243&stc=1&d=1342914228.gif

What is it that looks off to you?
 

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  • #3
I can not pinpoint anything. I am aware of the steps of the log diff for other examples on this weeks homework. Prof through it out to the class that this looks off. I cant see it. Should I look deeper?
 
  • #4
237
5
Looks perfectly fine to me.

Provided you understand the product rule, the chain rule, and the fact that ln(a^b) = b.ln(a), there isn't really anything mysterious here at all.
 
  • #5
ok thanks. Perhaps it was thrown out to check our grasp of your aforementioned rules. Thanks for the time.
 
  • #6
I threw this out last night, but it is still getting some pushback.

How does this look? Right or a little off? I placed the problem and solution link since I can not figure out how to import the image.

http://archives.math.utk.edu/visual....gdiff.1/a1.gif [Broken]
 
Last edited by a moderator:
  • #7
18
0
This is what I get when trying to look at the image:
The requested URL /visual....gdiff.1/a1.gif was not found on this server.
 
  • #9
The original problem is...

y = (2-x)^sqrt(x)
 
  • #10
18
0
And you want the derivative?
 
  • #11
yes, using logarithmic differentiation solve the problem.
 
  • #12
18
0
In general, when you have something like [tex]f(x)^{g(x)}[/tex] you can find the derivative like this, which should give you what you need to solve your specific problem where log is the natural logarithm:

[tex]y=f(x)^{g(x)}[/tex]
[tex]\log{y}=g(x)\log{f(x)}[/tex]
Implicitly differentiate both sides:
[tex]\frac{dy}{y}=g'(x)\log{f(x)}dx+\frac{g(x)f'(x)}{f(x)}dx[/tex]
[tex]y'=y\left(g'(x)\log{f(x)}+\frac{g(x)f'(x)}{f(x)} \right)[/tex]
[tex]y'=f(x)^{g(x)}\left(g'(x)\log{f(x)}+\frac{g(x)f'(x)}{f(x)}\right)[/tex]
 
  • #13
606
1
yes, using logarithmic differentiation solve the problem.



It looks fine.

DonAntonio
 
  • #14
237
5
Your efforts to spread this question to a larger audience have obviously been quashed, as it looks like it's been merged into your original thread!
I threw this out last night, but it is still getting some pushback.
What exactly do you mean by "some pushback"? Has your Prof said something? If so, what? Is he/she worried about the domain/range not being real etc?

Even the God Almighty that is Woolfy-Alfa agrees that this is the correct solution.
 
Last edited:

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