Question

Question #f6339

Answer 1

Please see the explanation below.

Explanation:

`V=sqrt(x^2+y^2+z^2)=(x^2+y^2+z^2)^(1/2)`

By taking the partial derivative w.r.t. `x`,

`V_x=1/cancel(2)(x^2+y^2+z^2)^(-1/2)cdot(cancel(2)x)=x/sqrt(x^2+y^2+z^2)`

By taking another partial derivative w.r.t. `x`,

`V_(x x)=(1cdot sqrt(x^2+y^2+z^2)-x cdot x/sqrt(x^2+y^2+z^2))/(sqrt(x^2+y^2+y^2))^2=(V-x^2/V)/V^2`

By multiplying the numerator and the denominator by `V`,

`Rightarrow V_(x x)=(V^2-x^2)/V^3`

By using symmetries,

`Rightarrow V_(y y)=(V^2-y^2)/V^3`

and

`Rightarrow V_(z z)=(V^2-z^2)/V^3`

Hence,

`V_(x x)+V_(y y)+V_(z z)=(V^2-x^2)/V^3+(V^2-y^2)/V^3+(V^2-z^2)/V^3`

`=(3V^2-(x^2+y^2+z^2))/V^3=(3V^2-V^2)/V^3=(2V^2)/V^3=2/V`

I hope that this was clear.