Find $f_x,f_y,f_x(-5,-1)$, and $f_y(2,-2)$ for the following equation.
$$f(x,y)=\sqrt{x^2+y^2}$$
$${\mathrm{f}}_{\mathrm{x}}=$$
(Type an exact answer, using radicals as needed.)
$$f_y=$$
(Type an exact answer, using radicals as needed.)
$$f_x(-5,-1)=$$
(Type an exact answer, using radicals as needed.)
$${\mathrm{f}}_{\mathrm{y}}(2,-2)=$$
(Type an exact answer, using radicals as needed.)

Step 1 ：The function given is $f(x,y)=\sqrt{x^2+y^2}$.

Step 2 ：The partial derivative of a function with respect to a variable is the derivative of the function with respect to that variable, treating all other variables as constants.

Step 3 ：To find $f_x$ and $f_y$, we need to take the derivative of $f(x,y)$ with respect to $x$ and $y$ respectively.

Step 4 ：The partial derivative of $f(x,y)$ with respect to $x$ is $f_x=\frac{x}{\sqrt{x^2+y^2}}$.

Step 5 ：The partial derivative of $f(x,y)$ with respect to $y$ is $f_y=\frac{y}{\sqrt{x^2+y^2}}$.

Step 6 ：To find $f_x(-5,-1)$ and $f_y(2,-2)$, we substitute these values into the partial derivatives we found.

Step 7 ：The value of $f_x$ at the point $(-5,-1)$ is $f_x(-5,-1)=\frac{-5}{\sqrt{26}}$.

Step 8 ：The value of $f_y$ at the point $(2,-2)$ is $f_y(2,-2)=\frac{-2}{\sqrt{8}}=-\frac{\sqrt{2}}{2}$.

Step 9 ：$\boxed{{\displaystyle f_x=\frac{x}{\sqrt{x^2+y^2}}}}$

Step 10 ：$\boxed{{\displaystyle f_y=\frac{y}{\sqrt{x^2+y^2}}}}$

Step 11 ：$\boxed{{\displaystyle f_x(-5,-1)=\frac{-5}{\sqrt{26}}}}$

Step 12 ：$\boxed{{\displaystyle f_y(2,-2)=-\frac{\sqrt{2}}{2}}}$