# Simple Partial Differentiation problem

1. Sep 13, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
If $z=x\ln(x+r)-r$ where $r^2=x^2+y^2$, prove that
$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{x+y}$$

2. Relevant equations

3. The attempt at a solution
Since $r^2=x^2+y^2$, $∂r/∂x=x/r$ and $∂r/∂y=y/r$.
Differentiating z w.r.t x partially,
$$\frac{∂z}{∂x}=\ln(x+r)+x\cdot\left(\frac{1}{x+r}\right)\cdot \left(1+\frac{∂r}{∂x}\right)-\frac{∂r}{∂x}$$
Using $∂r/∂x=x/r$
$$\frac{∂z}{∂x}=\ln(x+r)$$
$$\frac{∂^2z}{∂x^2}=\frac{1}{x+r}\cdot\left(1+\frac{x}{r}\right)=\frac{1}{r}$$

Differentiating z w.r.t y partially,
$$\frac{∂z}{∂y}=\frac{x}{x+r}\cdot \frac{∂r}{∂y}-\frac{∂r}{∂y}=\frac{∂r}{∂y}\cdot \frac{-r}{x+r}$$
Using $∂r/∂y=y/r$
$$\frac{∂z}{∂y}=\frac{-y}{x+r}$$
$$\frac{∂^2z}{∂y^2}=-\frac{(x+r)-y(∂r/∂y)}{(x+r)^2}=-\frac{rx+r^2-y^2}{r(x+r)^2}$$
Since $r^2-y^2=x^2$
$$\frac{∂^2z}{∂y^2}=-\frac{x}{r(x+r)}$$
Adding the second order derivatives,
$$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{r}-\frac{x}{r(x+r)}=\frac{1}{x+r}$$
Where did I go wrong?

Any help is appreciated. Thanks!

2. Sep 13, 2013

### pasmith

I don't think you have; I also get the same answer.

3. Sep 13, 2013

### Pranav-Arora

Thanks for the check pasmith!

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