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

Options

A. `-(cosx + sin x )` , for x `in (pi//4 , pi//2)`
B. cosx + sin x for x `in ( 0 , pi//4)`
C. `-(cos x + sinx )` , for `x in (0 , pi//4)`
D. `cos x - sin x ,` for `x in (pi//4 , pi//2)`

Answer

C

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):}`