Calculating Partial Derivative of Log(x^2+y^2) w/r/t x

  • Thread starter Somefantastik
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In summary, we discussed the process of taking the partial derivative of log(x^2 + y^2) with respect to x and the potential need for a change of base to log_e before performing the partial derivative. We also looked at the second partial derivative and how it can be used to show harmonicity using the Laplacian.
  • #1
Somefantastik
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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.
 
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  • #2
Somefantastik said:
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
Hi Somefantastik! :smile:
Somefantastik said:
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:
 

1. What is the formula for calculating the partial derivative of log(x^2+y^2) with respect to x?

The formula for calculating the partial derivative of log(x^2+y^2) with respect to x is 2x/(x^2+y^2).

2. How do you calculate the partial derivative of log(x^2+y^2) with respect to x?

To calculate the partial derivative of log(x^2+y^2) with respect to x, you need to use the quotient rule of differentiation and then simplify the resulting expression.

3. Can the partial derivative of log(x^2+y^2) with respect to x be simplified?

Yes, the partial derivative of log(x^2+y^2) with respect to x can be simplified to 2x/(x^2+y^2).

4. What does the partial derivative of log(x^2+y^2) with respect to x represent?

The partial derivative of log(x^2+y^2) with respect to x represents the instantaneous rate of change of the function with respect to x at a specific point on the graph.

5. Is the partial derivative of log(x^2+y^2) with respect to x always positive?

No, the partial derivative of log(x^2+y^2) with respect to x can be either positive or negative, depending on the values of x and y. This is because the function is a logarithm, and its derivative can be either positive or negative depending on the input values.

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