What is \(\frac{{{\rm{d}}\sqrt {1 - \sin 2{\rm{x}}} }}{{{\rm{dx}}}}\) equal to, where \(\frac{{\rm{\pi }}}{4} < {\rm{x}} < \frac{{\rm{\pi }}}{2}?\)

What is \(\frac{{{\rm{d}}\sqrt {1 - \sin 2{\rm{x}}} }}{{{\rm{dx}}}}\)�

1.	What is \(\frac{{{\rm{d}}\sqrt {1 - \sin 2{\rm{x}}} }}{{{\rm{dx}}}}\) equal to, where \(\frac{{\rm{\pi }}}{4} < {\rm{x}} < \frac{{\rm{\pi }}}{2}?\)
A.	cos x + sin x
B.	– (cos x + sin x)
C.	± (cos x + sin x)
D.	None of above
Answer» B. – (cos x + sin x)