Given,
$y=\sqrt{2 x-x^{2}}$
By differentiating both side w.r.t. 'x', we get
$\frac{d y}{d x} =\frac{1}{2 \sqrt{2 x-x^{2}}} \times(2-2 x) $
$=\frac{1-x}{\sqrt{2 x-x^{2}}}$
Here, $\frac{d y}{d x}$ is defined for
$2 x-x^{2}>\,0$
i.e. $ 0<\,x<\,2$
Now, $\frac{d y}{d x}>\,0$
When. $1-x>\,0$
$\left[\because \sqrt{2 x-x^{2}}\right.$ is always positive]
$\Rightarrow \,x<\,1$
$\therefore $ Given function increases in $0<\,x<\,1$.
And $\frac{d y}{d x}<\,0$
When,$ \int-x<0$
$\left[\because \sqrt{2 x-x^{2}}\right.$ is always positive]
$x>\,1$
$\therefore $ Given function decreases in $1<\,x<\,2$.