\(\displaystyle{f{{\left({x}\right)}}}={y}_{{{1}}}=\sqrt{{{x}}}-{x}^{{{2}}}\)

we have to find the derivative by using the power rule:

Power Rule:

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}^{{{n}}}={n}{\left({x}\right)}^{{{n}-{1}}}\)

Now

\(\displaystyle{y}_{{{1}}}={x}^{{{\frac{{{1}}}{{{2}}}}}}-{x}^{{{2}}}\)

\(\displaystyle{f}'{\left({x}\right)}={y}_{{{1}}}'={\frac{{{1}}}{{{2}}}}{x}^{{{\frac{{{1}}}{{{2}}}}-{1}}}-{2}{x}^{{{2}-{1}}}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}{x}^{{-{\frac{{{1}}}{{{2}}}}}}-{2}{x}\)

\(\displaystyle={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}-{2}{x}\)

Hence, \(\displaystyle{f}'{\left({x}\right)}={y}_{{{1}}}'={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}-{2}{x}\)

we have to find the derivative by using the power rule:

Power Rule:

\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}^{{{n}}}={n}{\left({x}\right)}^{{{n}-{1}}}\)

Now

\(\displaystyle{y}_{{{1}}}={x}^{{{\frac{{{1}}}{{{2}}}}}}-{x}^{{{2}}}\)

\(\displaystyle{f}'{\left({x}\right)}={y}_{{{1}}}'={\frac{{{1}}}{{{2}}}}{x}^{{{\frac{{{1}}}{{{2}}}}-{1}}}-{2}{x}^{{{2}-{1}}}\)

\(\displaystyle={\frac{{{1}}}{{{2}}}}{x}^{{-{\frac{{{1}}}{{{2}}}}}}-{2}{x}\)

\(\displaystyle={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}-{2}{x}\)

Hence, \(\displaystyle{f}'{\left({x}\right)}={y}_{{{1}}}'={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}-{2}{x}\)