Answer
$f'(x)=\dfrac{x-2}{\sqrt{x^2-4x+2}}$
Work Step by Step
$u=(x^2-4x+2)$; $\dfrac{du}{dx}=(2x-4)$
$f(u)=\sqrt{u}; \dfrac{d}{du}f(u)=\dfrac{1}{2\sqrt{u}}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}$ $=(\dfrac{1}{2\sqrt{(x^2-4x+2)}})(2x-4)$ $=\dfrac{x-2}{\sqrt{x^2-4x+2}}$
