If y=cos−1⁡2x1+x2,−1

Substitute x=tan⁡θ⇒θ=tan−1⁡x y=cos−1⁡2tan⁡θ1+tan2⁡θ⇒ y=cos−1⁡(sin⁡2θ) ∵sin⁡2θ=2tan⁡θ1+tan2⁡θ⇒ y=cos−1⁡cos⁡π2−2θ ∵sin⁡2θ=cos⁡π2−2θ⇒ y=π2−2θ⇒y=π2−2tan−1⁡x∵θ=tan−1⁡xOn differentiating  both sides  vv.r.t.x, we getdydx=0−21+x2⇒ dydx=−21+x2 ∵ddxtan−1⁡x=11+x2