Derivative of sin−1⁡1x+1 with respect to x is

Let   y=sin−1⁡1x+1Let x=tan2⁡θ⇒θ=tan−1⁡xy=sin−1⁡11+tan2⁡θ=sin−1⁡1sec⁡θy=sin−1⁡(cos⁡θ)+sin−1⁡{sin⁡(π/2−θ)}y=π2−θ=π2−tan−1⁡xNow,  differentiating  w.r.t .x, we  getdydx=0−11+x⋅ddx(x)⇒dydx=−12x(1+x)