`(d)/(dx)(cos (x/2))=`
`(A)` `(1)/(2)sin(x)/(2)` , ` (B)` `-(1)/(2)cos(x)/(2)`
` (C)` `-(1)/(2)sin(x)/(2)`, ` (D)` `-sin(x)/(4)`

d/(dx)[cos(1-x^2)^2]=

(d)/(dx)cos^(-1)(sin x)=

d/(dx) (sin x+ cos x)

(d)/(dx)(sin^(2)x+cos^(2)x) =

If int sqrt(1+sec x)dx=2sin^(-1)[sqrt(2)f(x)]+c then (d)/(dx)[f(x)] is equal to (A)cos(x)/(2)(B)-(1)/(2)cos(x)/(2)(C)(1)/(2)cos(x)/(2)(D)(1)/(2)sin(x)/(2)

(d)/(dx)[(sin x)/(1+cos x)]=

(d)/(dx)[sin^(-1){cos(x^(2)-2)}]=

(d) / (dx) (cos ^ (2) (tan ^ (- 1) (sin (cot ^ (- 1) x))))

(d)/(dx)[3(sin^(2)x+cos^(2)x)]=

(d)/(dx)cos x^(0) = (A) -sin x^(0) (B) -(pi)/(180)sin x^(0) ( C) (pi)/(180)sin x^(0) (D) (2 pi)/(180)sin x^(0)