Question

If x < 2, then write the value of $\frac{d}{dx}\left(\sqrt{x^2-4x+4}\right)$.

Answer 1

$\frac{d}{dx}\left(\sqrt{x^2-4x+4}\right)=\frac{d}{dx}\left(\sqrt{(x-2{)}^2}\right)$

$=\frac{d}{dx}(|x-2|)$

$=\frac{d}{dx}(-(x-2\left)\right)(\because x<2)$

$=-1$