elementary partial derivative proof.

Question

If $z=e^{k(r-x)} r^2=x^2+y^2,\\$ show that: $\\\biggr(\frac{\partial{z}}{\partial{x}}\biggr)^2 +\biggr(\frac{\partial{z}}{\partial{y}}\biggr)^2+2zk\frac{\partial{z}}{\partial{x}}=0\\$

$My\;attempt :\\ r=\sqrt{x^2+y^2};$

$\frac{\partial{z}}{\partial{r}}=ke^{k(r-x)} $ $\frac{\partial{z}} {\partial{x}}=-ke^{k(r-x)};$

$\frac{\partial{r}}{\partial{x}}$=$\frac{x} {\sqrt{x^2+y^2} }$

$\frac{\partial{r}}{\partial{y}}$=$\frac{y}{\sqrt{x^2+y^2}}$ ; $\frac{\partial{z}}{\partial{y}}$=$\frac{\partial{z}}{\partial{r}}\cdot\frac{\partial{r}}{\partial{y}}$=$ke^{k(r-x)}\cdot\frac{y}{\sqrt{x^2+y^2}}\Biggr|\frac{\partial{z}}{\partial{x}}=\frac{\partial{z}}{\partial{r}}\cdot\frac{\partial{r}}{\partial{x}}={-ke^{k(r-x)}}\cdot\frac{x} {\sqrt{x^2+y^2}}$

$\biggr(\frac{\partial{z}}{\partial{x}}\biggr)^2+$$\biggr(\frac{\partial{z}}{\partial{y}}\biggr)^2$=$\Biggr[\frac{-kxe^{k(r-x)}}{\sqrt{x^2+y^2}}$ $\Biggr]^2+\Biggr[\frac{kye^{k(r-x)}}{\sqrt{x^2+y^2}}\Biggr]^2\\k^2e^{2k(r-x)}\Biggr[\frac{x^2} {x^2+y^2}+\frac{y^2}{x^2+y^2}\Biggr]=k^2e^{2k(r-x)}$ $\\2zk\frac{\partial{z}} {\partial{x}}=2e^{k(r-x)}(k)(-ke^{k(r-x})=-2k^2e^{2k(r-x)}$$\biggr(\frac{\partial{z}}{\partial{x}}\biggr)^2 +\biggr(\frac{\partial{z}}{\partial{y}}\biggr)^2$+$2zk\frac{\partial{z}}{\partial{x}}=-k^2e^{2k(r-x)}$

Where am I getting wrong? Pardon my typesetting

Answer 1

$z = e^{k (r - x)}, r^2 = x^2 + y^2$. Then $$z = e^{k (\sqrt{x^2 + y^2}- x)}$$ $$\frac{\partial z}{\partial x} = \frac{\partial e^{k X}}{\partial X} \frac{\partial X}{\partial x}$$ with $X= \sqrt{x^2 + y^2}- x$ and $\frac{\partial e^{k X}}{\partial X} = k e^{k X}$ while $\frac{\partial X}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} - 1$. Similarly, $\frac{\partial X}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}}$.

Putting these results together $$\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = k^2 z^2 \left[ \left( \frac{x}{\sqrt{x^2 + y^2}} - 1\right)^2 + \left( \frac{y}{\sqrt{x^2 + y^2}} \right)^2 \right]$$

Expanding gives the rhs as $$k^2 z^2 \left[ 1 - \frac{2x}{\sqrt{x^2 + y^2}} + 1 \right] = - 2 k^2 z^2 \left[ \frac{x}{\sqrt{x^2 + y^2}} - 1 \right].$$

The result follows when you use the result $\partial z/\partial x = k z \left( \frac{x}{\sqrt{x^2 + y^2}} - 1 \right)$.

Answer 2

I am almost certain that $z(x,y)=z^*(r(x,y),x)$ is considered a function of $x$ and $y$. Here $r$ is merely a placeholder. Following this, you should use $$ \frac{\partial z}{\partial r} = \frac{\partial z^*}{\partial x} + \frac{\partial z^*}{\partial r} \frac{\partial r}{\partial x} $$