If `f(x)=sqrt(1-sin2x)` , then `f^(prime)(x)` is equal to

(a)`-(cosx+sinx)`,for `x in (pi/4,pi/2)`
(b)`cosx+sinx` ,for `x in (0,pi/4)`
(c)`-(cosx+sinx)`,for `x in (0,pi/4)`
(d)`cosx-sinx` ,for `x in (pi/4,pi/2)`

Explanation

`f(x)=sqrt(1-sin 2x)=sqrt((cos x- sin x)^(2))`
`=|cos x - sin x|`
`={(cos x- sin x","," for "0 lexlepi//4),(-(cos x - sin x)","," for "pi//4ltxlepi//2):}`
`therefore" "f'(x)={{:(-(cos x + sin x)","," for "0ltxltpi//4),((cos x + sin x)","," for "pi//4ltxltpi//2):}`