Question

If $u=(x^2+y^2+z^2{)}^{1/2}$ then prove that $u(\frac{d^{2}u}{d{x}^2}+\frac{d^{2}u}{d{y}^2}+\frac{d^{2}u}{s{z}^2})=2$.

Answer 1

$u=\sqrt{x^2+y^2+z^2}$

$\frac{du}{dx}=\frac{1}{2\sqrt{x^2+y^2+z^2}}\left(2x\right)=\frac{x}{\sqrt{x^2+y^2+z^2}}$

$\frac{d^{2}u}{d{x}^2}=\frac{\sqrt{x^2+y^2+z^2}\frac{d}{dx}\left(x\right)-x\frac{d}{dx}\left(\sqrt{x^2+y^2+z^2}\right)}{(\sqrt{x^2+y^2+z^2}{)}^2}$

$=\frac{\sqrt{x^2+y^2+z^2}-x\frac{1}{2\sqrt{x^2+y^2+z^2}}\left(2x\right)}{x^2+y^2+z^2}$

$=\frac{x^2+y^2+z^2-x^2}{(x^2+y^2+z^2)\sqrt{x^2+y^2+z^2}}$

$=\frac{y^2+z^2}{(x^2+y^2+z^2{)}^{3/2}}$

Similarly,

$\frac{d^{2}u}{d{y}^2}=\frac{x^2+z^2}{(x^2+y^2+z^2{)}^{3/2}}$,

$\frac{d^{2}u}{d{z}^2}=\frac{x^2+y^2}{(x^2+y^2+z^2{)}^{3/2}}$

$u(\frac{d^{2}u}{d{x}^2}+\frac{d^{2}u}{d{y}^2}+\frac{d^{2}u}{d{z}^2})=\sqrt{x^2+y^2+z^2}$$\left(\frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^{3/2}}\right)$.

$=\frac{(x^2+y^2+z^2{)}^{3/2}.2}{(x^2+y^2+z^2{)}^{3/2}}=2$.