D/dx log(x^2+y^2)

  • #1
I'm trying to get

[tex] \frac{\partial}{\partial x} log(x^{2} + y^{2}) [/tex]

let z = x2+y2

Do I need to do a change of base to go from log10z to logez before I can do the partial w.r.t. x?

That would make it

[tex] \frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)[/tex]

Does this look right?

Then

[tex]\frac{\partial^{2}}{\partial x^{2}}(log(x^{2}+y^{2})) = - \frac{1.72x^{2}}{(x^{2} + y^{2})^{2}} + 0.86(x^{2}+y^{2}) [/tex]

?? that doesn't look right.
 

Answers and Replies

  • #2
tiny-tim
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Do I need to do a change of base to go from log10z to logez before I can do the partial w.r.t. x?

That would make it

[tex] \frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)[/tex]

Hi Somefantastik! :smile:

Yes, that's fine …

logab = [tex]\frac{log_eb}{log_a}[/tex]

and of course logab = 1/logba :wink:
[tex]\frac{\partial^{2}}{\partial x^{2}}(log(x^{2}+y^{2})) = - \frac{1.72x^{2}}{(x^{2} + y^{2})^{2}} + 0.86(x^{2}+y^{2}) [/tex]

?? that doesn't look right.

ha ha!

how about [tex]\frac{0.86}{(x^{2}+y^{2})} [/tex] ? :smile:
(and then of course combine both fractions into one)
 
  • #3
Hello tiny-tim, from the sunny south [usa] :)

That gives -0.86 /(x2 + y2)

It just seems like a weird number to me.

That's going to be the same number for the partial with respect to y, isn't it?

They should cancel out; I'm trying to show that this is a harmonic using the laplacian.
 
  • #4
tiny-tim
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Homework Helper
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Hi Somefantastik! :smile:
That's going to be the same number for the partial with respect to y, isn't it?

They should cancel out; I'm trying to show that this is a harmonic using the laplacian.

Yup …

[tex]\left(\frac{\partial^{2}}{\partial x^{2}}\ +\ \frac{\partial^{2}}{\partial y^{2}}\right)(log(x^{2}+y^{2})) = - \frac{1.72(x^{2}\,+\,y^2)}{(x^{2} + y^{2})^{2}}\ +\ 0.86/(x^{2}+y^{2})\ +\ + 0.86/(x^{2}+y^{2})\ =\ 0\ [/tex] :wink:
 

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