Can anyone differentiate this for me?

[roars]

Can anyone differentiate this for me?

Given that V= 1/(z^m), z = (a^2+b^2+c^2)^(1/2), show that for (z!=0)

d2V/da2 + d2V/db2 + d2V/dc2 = m(m-1)/(z^(m+2))

Any help would be appreciated!

Cheers Ruairi

[krtxmrtz]

dV/da=(dV/dz)(dz/da)=(-mz^(-m-1))(a/(a^2+b^2+c^2)^(1/2))=
(-mz^(-m-1))(a/z)=-maz^(-m-2)

d2V/da2=d/da(-maz^(-m-2))=-mz^(-m-2)+am(m+2)z^(-m-3)(dz/da)
=-mz^(-m-2)+am(m+2)z^(-m-3)(a/z)
=-mz^(-m-2)+m(m+2)(a^2)(z^(-m-4))

For the b and c terms the result is the same where a above must be replaced by b and c respectively. Adding them up:

d2V/da2+d2V/db2+d2V/dc2=
-3mz^(-m-2)+(a^2+b^2+c^2)m(m+2)z^(-m-4)=
-3mz^(-m-2)+(z^2)m(m+2)z^(-m-4)=
-3mz^(-m-2)+m(m+2)z^(-m-2)=
(m^2+2m-3m)z^(-m-2)=m(m-1)z^(m-1)

Yes, I know, it looks messy but I hope there aren't any typos.

[krtxmrtz]

Damn!, there WAS a mistake at the very end of the demonstration. The last line should be:

... =(m^2+2m-3m)z^(-m-2)=m(m-1)z^(m+2)