Question

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

Answer 1

$$\begin{gathered} \text{Given: x<2} \\ \text{∴ 2-x>0} \\ \frac{d}{dx}\left(\sqrt{x^2-4x+4}\right) \\ =\frac{d}{dx}\left(\sqrt{{\left(2-x\right)}^2}\right) \\ =\frac{d}{dx}\left(2-x\right)\text{(∵ 2-x>0)} \\ =0-1 \\ =-1 \end{gathered}$$