Answer
$\dfrac {2\left( x^{2}-x+1\right) }{x^{2}-2x+2} $
Work Step by Step
$\dfrac {d}{dx}\left( 2x\sqrt {x^{2}-2x+2}\right) =\left( \dfrac {d}{dx}\left( 2x\right) \right) \times \sqrt {x^{2}-2x+2}+\left( \dfrac {d}{dx}\sqrt {x^{2}-2x+2}\right) \times 2x=2\sqrt {x^{2}-2x+2}+\dfrac {1}{2}\times \dfrac {\left( 2x-2\right) }{\sqrt {x^{2}-2x+2}}\times 2x=\dfrac {2\left( x^{2}-2x+2+x-1\right) }{x^{2}-2x+2}=\dfrac {2\left( x^{2}-x+1\right) }{x^{2}-2x+2} $
