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

I'm trying to get

$$\frac{\partial}{\partial x} log(x^{2} + y^{2})$$

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

$$\frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)$$

Does this look right?

Then

$$\frac{\partial^{2}}{\partial x^{2}}(log(x^{2}+y^{2})) = - \frac{1.72x^{2}}{(x^{2} + y^{2})^{2}} + 0.86(x^{2}+y^{2})$$

?? that doesn't look right.

tiny-tim
Homework Helper
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

$$\frac{\partial}{\partial x} log(x^{2} + y^{2}) = \frac{1}{x^{2}+y^{2}} \ (2x) \ (log_{10}e)$$

Hi Somefantastik!

Yes, that's fine …

logab = $$\frac{log_eb}{log_a}$$

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

?? that doesn't look right.

ha ha!

how about $$\frac{0.86}{(x^{2}+y^{2})}$$ ?
(and then of course combine both fractions into one)

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.

tiny-tim
$$\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\$$