Question

differentiation sinx

Answer 1

y = sin x

dy / dx = Lim  Δx -> 0  Δy/Δx
        =  Lim Δx -> 0,   [ Sin (x + Δx) - Sin x ] / (x+Δx  - x)
         =  2 Cos ( x + Δx/2 ) Sin Δx/2    / Δx
       = lim Δx -> 0, Cos ( x + Δx/2 )  *  Sin (Δx/2) /  (Δx/2)

as Δx -> 0, Δx/2 -> 0,  then  Sin (Δx/2) / (Δx/2)  =  1

dy/dx =  lim Δx/2 ->0  Cos (x + Δx/2) = Cos x