If\ f(x)=\sqrt{1-sin(2x)},f'(x) equals\\\ \\ a)-(cos x + sin x) for\ x\in(0,\pi/4)\\b)(cos x + sin x) for\ x\in(\pi/4,\pi/2)\\c)-(cos x + sin x) for\ x\in(\pi/4,\pi/2)\\d)(cos x + sin x) for\ x\in(0,\pi/4)\\
1Please let me know the flaw in my solution..
f'(x)=\frac{|cosx-sinx|}{cosx-sinx}\times -(sinx+cosx)\\\ \\ If\ x\in (\pi/4,\pi/2),tanx>1\ or\ cosx-sinx<0\\\ \\ so\ for\ x\in (\pi/4,\pi/2),f'(x)=cosx+sinx
11It could be written as
\sqrt{sin^{2}x+cos^{2}x-2sinx cosx}
which is
|sinx-cosx|
now 2 cases arise
when sinx >cosx(∩/4,∩/2)................sinx -cosx
and cos x >sinx (0,∩/4)....................cosx-sinx
so f'(x) is cosx+sinx(∩/4,∩/2)
and -(sinx+cosx)(0,∩/4)
so answer is a and b
if its not matching then the answer given must ewrong
