So $f$ is the function defined by $f(x)=\sqrt[\phantom{2}]{2x-x^2}$

$($a$)$ Its field of definition is

$R.$

$0,2.$

$.$

$($b$)$ Its derivative is

$f^{\text{'}}(x)=\frac{1-x}{\sqrt[\phantom{2}]{2x-x^2}}$

$f^{\text{'}}(x)=\frac{x-1}{2\sqrt[\phantom{2}]{2x-x^2}}$

$f^{\text{'}}(x)=\frac{1}{2\sqrt[\phantom{2}]{2x-x^2}}$