Partial Derivatives Homework: Find Sum of Second Partials

[wilcofan3]

Homework Statement

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


Find $$\frac {\partial^2 u} {\partial x^2} + \frac {\partial^2 u} {\partial y^2} + \frac {\partial^2 u} {\partial z^2}$$

Homework Equations

The Attempt at a Solution

$$\frac {\partial^2 u} {\partial x^2} = -(x^2 + y^2 +z^2)^\frac {-3} {2} + 3x^2(x^2 + y^2 + z^2)^\frac {-5} {2}$$

$$\frac {\partial^2 u} {\partial y^2} = -(x^2 + y^2 +z^2)^\frac {-3} {2} + 3y^2(x^2 + y^2 + z^2)^\frac {-5} {2}$$

$$\frac {\partial^2 u} {\partial z^2} = -(x^2 + y^2 +z^2)^\frac {-3} {2} + 3z^2(x^2 + y^2 + z^2)^\frac {-5} {2}$$

So I think all the partials are right, but I feel like I'm getting a crazy answer when I add them together.

$$3x^2 + 3y^2 + 3z^2(x^2 + y^2 + z^2)^\frac {-5} {2} -3(x^2 + y^2 + z^2)^\frac {-3} {2}$$

Is this right?

 

[Dick]

Why do you think that's crazy? It looks correct to me. But you can express the answer in a much simpler form.

 

[rootX]

here's the MATLAB quick code for the first step - finding d^2/dx^2:

>> syms u; syms y; syms z;
>> u = 1/sqrt(x^2+y^2+z^2)

u =

1/(x^2+y^2+z^2)^(1/2)


>> diff(u,x)

ans =

-1/(x^2+y^2+z^2)^(3/2)*x


>> diff(ans,x)

ans =

3/(x^2+y^2+z^2)^(5/2)*x^2-1/(x^2+y^2+z^2)^(3/2)