Hessian of a function Given this equation: $f(x,y,z) = \sqrt{1+x^2+y^2+z^2}$
I tried to calculate the Hessian --> for example $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ The second derivativ respect to x is hard to calculate for me: I tried the product rule: $x*(1+x^2+y^2+z^2)^{-1/2}$.
Then i get: $\frac{1}{1+x^2+y^2+z^2} - \frac{x^2}{(1+x^2+y^2+z^2)^{3/2}}$
But this is wrong.
Could anyone help?
 A: What you get it's quite correct, because $$\frac{\partial f}{\partial x^2}=\frac{1}{\sqrt{1+x^2+y^2+z^2}}-\frac{x^2}{(1+x^2+y^2+z^2)^{3/2}}$$ 
And making some algebra you get $$\frac{\partial f}{\partial x^2}= \frac{1+y^2+z^2}{(1+x^2+y^2+z^2)^{3/2}}$$
I think that the other may be similar to this
A: Indeed $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ which you can write as $\frac{\partial f}{\partial x} = x(1+x^2+y^2+z^2)^{-\frac{1}{2}}$ so that $$\frac{\partial^2 f}{\partial x^2} = (1+x^2+y^2+z^2)^{-\frac{1}{2}} -\frac{1}{2} 2x^2 (1+x^2+y^2+z^2)^{-\frac{1}{2}-1} $$ $$ = (1+x^2+y^2+z^2)^{-\frac{1}{2}} - x^2 (1+x^2+y^2+z^2)^{-\frac{3}{2}}$$ $$ = (1+x^2+y^2) / \sqrt{1+x^2+y^2+z^2}^3$$ by putting $1/ \sqrt{1+x^2+y^2+z^2}$ in factor, and $$\frac{\partial^2 f}{\partial x \partial y} = xy (1+x^2+y^2+z^2)^{-\frac{3}{2}}.$$
Do you see now what you don't need to perform any other calculation to get all other partial derivatives needed to calculate the Hessian matrix ?
Note that I just used this : if $f(t)=g(t)^{\alpha}$ then $f'(t)=\alpha g'(t)g(t)^{\alpha-1}$ if $g$ is differentiable in one variable $t$
