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Answer 1 · 2017-04-19T10:16:08

Please see the explanation below.

#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.