Aus Online Mathematik Brückenkurs 2

If we write \displaystyle \sqrt{x} in power form \displaystyle x^{{1}/{2}\;}, we see that the square root is a function having the appearance of \displaystyle x^{n} and its derivative is therefore equal to

\displaystyle {f}'\left( x \right)=\frac{d}{dx}\sqrt{x}=\frac{d}{dx}x^{{1}/{2}\;}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}

The answer can also be written as

\displaystyle {f}'\left( x \right)=\frac{1}{2\sqrt{x}}

because \displaystyle x^{-\frac{1}{2}}=\left( x^{\frac{1}{2}} \right)^{-1}=\left( \sqrt{x} \right)^{-1}=\frac{1}{\sqrt{x}}