Finding $\frac{\partial}{\partial x}\left(x^2+y^2+z^2\right)^{-1}$

Question

Find the derivative with respect to x for the following function:

$\left(x^2+y^2+z^2\right)^{-1}$

Answer 1

$$f= \left(x^2+y^2+z^2\right)^{-1}$$

$$f_x= -1(x^2+y^2+z^2)^{-2}\cdot 2x$$

$$f_x= \frac{-2x}{(x^2+y^2+z^2)^{2}}$$

Answer 2

$$ f=(u(x))^n\\f'=n(u(x))^{n-1}(u(x))'\\$$so$$f=u^{-1}\\f'=-1u^{-2}u' $$

Answer 3

HINT:

Let $f(x)=1/(x^2+a)$. Then $f'(x)=(-2x)/(x^2+a)^2$.