Question

If $f\left(x\right)=\sqrt{1-\sin 2x},$ then $f^{\prime }\left(x\right)$ is equal to

• A. $-(\cos x+\sin x)$, for $x\in (\pi /4,\pi /2)$
• B. $\cos x+\sin x,$ for $x\in (0,\pi /4)$
• C. $-(\cos x+\sin x)$, for $x\in (0,\pi /4)$
• D. $(\cos x-\sin x)$, for $x\in (\pi /4,\pi /2)$

Answer 1

Answer: (C)

$-(\cos x+\sin x)$, for $x\in (0,\pi /4)$

$f\left(x\right)=\sqrt{1-\sin 2x}=\sqrt{(\cos x-\sin x{)}^2}$

$=|\cos x-\sin x|$

$=\left\{\begin{array}{cc}\cos x-\sin x,& \text{for}0\le x\le \pi /4\\ -(\cos x-\sin x),& \text{for}\pi /4<x\le \pi /2\end{array}\right\}$

$\therefore f^{\prime }\left(x\right)=\left\{\begin{array}{cc}-(\cos x+\sin x),& \text{for}0<x<\pi /4\\ (\cos x+\sin x),& \text{for}\pi /4<x<\pi /2\end{array}\right\}$