Question

if z=xlog(x+r)-r where r²=x²+y² prove that d²z/dx²+d²z/dy²=1/x+y,d³z/dx³=-x/r³​

Answer 1

(1) $r^2=x^2+y^2$

$rr_x=x\\rr_y=y$

(2) $s=x+r$

$s_x=1+r_x=1+\dfrac{x}{r}=\dfrac{s}{r}\to \dfrac{s_x}{s}=\dfrac{1}{r}$

$s_y=r_y$

(3)

$z=x\log s -r\\\\z_x=\log s + \dfrac{s_x}{s}\cdot x-r_x\\\\z_x=\log s+\dfrac{x}{r}-\dfrac{x}{r}\\\\z_x=\log s\\\\z_{xx}=\dfrac{s_x}{s}=\dfrac{1}{r}\\ \\\\z_{xxx}=-\dfrac{r_x}{r^2}=\dfrac{\frac{x}{r}}{r^2}\\ \\\\\boxed{z_{xxx}=-\dfrac{x}{r^3}}\\\\-------------------------\\\\z_y=\dfrac{s_y}{s}\cdot x-r_y=\dfrac{r_y}{s}\cdot x-r_y=r_y(\dfrac{x}{s}-1)\\ \\\\z_y=r_y\cdot \dfrac{x-s}{s}=\dfrac{y}{r}\cdot \dfrac{-r}{s}\\\\z_y=-\dfrac{y}{s}$

$z_{yy}=-\dfrac{s-ys_y}{s^2}=-\dfrac{s-yr_y}{s^2}\\ \\\\z_{yy}=-\dfrac{s-\frac{y^2}{r}}{s^2}\\ \\ \\z_{yy}=-\dfrac{xr+r^2-y^2}{rs^2}\\ \\ \\z_{yy}=-\dfrac{xr+x^2}{rs^2}=-\dfrac{x}{rs}\\ \\ \\\\z_{xx}+z_{yy}=\dfrac{1}{r}-\dfrac{x}{rs}=\dfrac{s-x}{rs}\\ \\ \\z_{xx}+z_{yy}=\dfrac{1}{s}\\ \\ \\\boxed{z_{xx}+z_{yy}=\dfrac{1}{x+r}}$