Calculation of a partial derivative I have to find the $\frac{\partial}{\partial x}\left( f(x,y)\right)=\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$.
1) Thinking of that as $\frac{\partial}{\partial x}\left( (x^2+y^2)^{-\frac{1}{2}}\right)$ and $\frac{\partial}{\partial a}x^a = ax^{a-1}$ I get $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right) = -\frac{x}{\Vert (x,y) \Vert^3}$.
2) Thinking of that as $\frac{\partial}{\partial x}\left( (\sqrt{x^2+y^2})^{-1}\right)$ and $\frac{\partial}{\partial a}\sqrt{a}=\frac{1}{2\sqrt{a}}$ I get $x \cdot (x^2+y^2)$.
3) Thinking of that as $\frac{\partial}{\partial x}\left( \frac{1}{\sqrt{x^2+y^2}}\right)$ and $\frac{\partial}{\partial a}\frac{1}{a}=-\frac{1}{a^2}$ I get $-\frac{2x}{x^2+y^2}$.
Why 2) and 3) are not correct?

In your calculations you should use these 3 formulas:


*

*$\frac{\partial}{\partial x}\frac{1}{a(x,y)} = \frac{-1}{a^2(x,y)} \cdot \frac{\partial}{\partial x} a(x,y)$

*$\frac{\partial}{\partial x}\sqrt{b(x,y)} = \frac{1}{2\sqrt{b(x,y)}} \cdot \frac{\partial}{\partial x} b(x,y)$

*$\frac{\partial}{\partial x}(x^2+c(y)) = 2x$


In point 2:


*

*$c(y)=y^2$

*$a(x,y) = x^2 + c(y)$

*$b(x,y) = \frac{1}{a(x,y)}$

*Compute $\frac{\partial}{\partial x} \sqrt{b(x,y)}$


In point 3 :


*

*$c(y)=y^2$

*$b(x,y) = x^2+c(y) $

*$a(x,y) = \sqrt{b(x,y)} $

*Compute $\frac{\partial}{\partial x} \frac{1}{a(x,y)}$

