Question

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

Answer 1

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

$u=log(x^2+y^2+z^2{)}^{\frac{1}{2}}$

$u=\frac{1}{2}log(x^2+y^2+z^2)$

$\Rightarrow 2u=log(x^2+y^2+z^2)$

Differentiating w.r.t. 'x' on both sides, we get

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

$\Rightarrow \frac{du}{dx}=\frac{x}{x^2+y^2+z^2}$

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

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

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

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

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

$(x^2+y^2+z^2)(\frac{d^{2}u}{d{x}^2}+\frac{d^{2}u}{d{y}^2}+\frac{d^{2}u}{d{z}^2})=(x^2+y^2+z^2)\left[\frac{x^2+y^2+z^2}{(x^2+y^2+z^2{)}^2}\right]$

$=\frac{(x^2+y^2+z^2)}{(x^2+y^2+z^2{)}^2}=1$.