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mrm0607
Senior Member
Joined: 01 Apr 2009
Posts: 79
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 Posted: Tue, 3 Nov 2009 02:06:56 UTC Post subject: differentiation log functions q2 |
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Hi,
I need help to solve the following question.
y = sqrt(x^x), find y'
I am having hard time to make any progress.
This is all I know I can use:
ln(y) = ln(sqrt(x^x))
= (1/2)ln(x^x)
y'/y = (0)(ln(x^x)) + (1/2)d/dx(ln(x^x))
= (1/2)(1/(x^x).x.x^(x-1)
= 1/2
I think I messed up.
I would appreciate help.
Thank you. |
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KArts
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Joined: 15 Oct 2009
Posts: 28
Location: Milwaukee
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| Posted: Tue, 3 Nov 2009 04:19:37 UTC Post subject: |
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I think you can do...
lny = ln((x^x)^(0.5))
lny = (x*0.5)ln(x)
and then the right side is just a product rule. Be careful though because the ln (x) might be ln |x|. |
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math4ever
Member of the 'S.O.S. Math' Hall of Fame
Joined: 02 Jan 2005
Posts: 323
Location: Austin, Texas
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| Posted: Tue, 3 Nov 2009 04:23:07 UTC Post subject: |
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Since y = sqrt(x^x)
then ln y = .5x ln(x)
So y'(1 / y) = .5ln x + .5x(1/x) = .5lnx + .5 = .5(1+lnx)
and y ' = .5y(1+ ln x )
but y = sqrt(x^x), so we get:
y ' = .5 sqrt(x^x)(1+ln x)
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Math and alcohol don't mix....
Don't drink and derive. |
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math4ever
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| Posted: Tue, 3 Nov 2009 04:25:33 UTC Post subject: |
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x has to be positive for y to be well defined so we don't need |x| in the ln
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Math and alcohol don't mix....
Don't drink and derive. |
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KArts
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Joined: 15 Oct 2009
Posts: 28
Location: Milwaukee
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| Posted: Tue, 3 Nov 2009 04:32:58 UTC Post subject: |
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| Quote: | | x has to be positive for y to be well defined so we don't need |x| in the ln |
right-o |
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mrm0607
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Joined: 01 Apr 2009
Posts: 79
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| Posted: Wed, 4 Nov 2009 18:19:03 UTC Post subject: |
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Thank you.
I made a mistake in taking ln of right side.
| Quote: | x has to be positive for y to be well defined so we don't need |x| in the ln
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Could you please explain the above comment a little more with example.
Thanks. |
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KArts
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Posts: 28
Location: Milwaukee
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| Posted: Thu, 5 Nov 2009 02:34:06 UTC Post subject: |
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| Quote: |
Could you please explain the above comment a little more with example.
Thanks. |
Sure. The original function is y = √(x^x). Basically, you cannot take the square root of a negative number (and be solely in the set of real numbers). If x is negative and it is taken to a negative power, then it will be negative. Also, you cannot take 0^0. Therefore, only positive x values are in the function's domain.
Likewise, the log of a negative number is not a real solution (see: http://betterexplained.com/articles/demystifying-the-natural-logarithm-ln/). I didn't really take the time to analyze the original domain, so I was just cautioning you that sometimes you need to use absolute values after using ln transformations.
Example: y = (x^2)^(x^2)
ln(y) = ln (x^2)^(x^2)
ln(y) = (x^2)*ln(x^2)
ln(y) = 2(x^2)*ln(x)
But you have to specify the ln(x) only for positive x values, so the best you can do is...
ln(y) = 2(x^2)*ln|x|
math4ever was saying that since only positive x values are in the original domain of your function, you do not need to add the absolute value sign. The x values are already positive. |
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aswoods
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Joined: 23 Feb 2009
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Location: Adelaide, Australia
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| Posted: Thu, 5 Nov 2009 03:16:16 UTC Post subject: |
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| Quote: | | If x is negative and it is taken to a negative power, then it will be negative. |
Not necessarily! (-2)^(-2) = 1/4. The problem is that x^x is an imaginary or complex number if x is negative and not an integer. |
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KArts
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Posts: 28
Location: Milwaukee
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| Posted: Thu, 5 Nov 2009 03:39:48 UTC Post subject: |
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| Quote: | | Not necessarily! (-2)^(-2) = 1/4. The problem is that x^x is an imaginary or complex number if x is negative and not an integer. |
Hmm. Right you are as well. Darn you math for being so tricky!
Also, I should append that 0^0 isn't necessarily undefined, but can be thought of as 1 under many cases. And that opens up a whole different crazy debate that should probably be avoided.
Thanks for the check though aswoods. I am still learning myself, and so will be liable to make mistakes sometimes.
Cheers |
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KArts
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Posts: 28
Location: Milwaukee
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| Posted: Thu, 5 Nov 2009 03:51:14 UTC Post subject: |
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What about (-0.4)^(-0.4) though?
(-0.4)^(-0.4) = 1.44269906 etc. |
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aswoods
S.O.S. Oldtimer
Joined: 23 Feb 2009
Posts: 289
Location: Adelaide, Australia
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| Posted: Thu, 5 Nov 2009 04:58:29 UTC Post subject: |
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You are interpreting the exponent -0.4 as -2/5, and applying it as ((x^2)^(1/5))^(-1), so the result is positive.
But then if the exponent were just a little bit smaller: -2000001/5000001 it would yield a negative result. If the denominator were 5000000, it would give a complex result. There is no limiting behaviour, and so irrational exponents can't be assigned a value. To avoid this, use Euler's formula.
First, 
So 
Therefore (-0.4)^(-0.4) = 0.4458187887 - 1.3720891466 i |
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KArts
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Joined: 15 Oct 2009
Posts: 28
Location: Milwaukee
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| Posted: Thu, 5 Nov 2009 06:03:03 UTC Post subject: |
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Thank you for taking the time to go over that. It took me about 15 minutes to figure out what that meant, but that is pretty good time when working with complex numbers. Good explanation.
Kevin |
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