Find dydxin the following questions:
y=sin−1(2x√1−x2), −1√2<x<1√2
Let sin−1(2x√1−x2)⇒y=sin−1(2sin θ√1−sin2θ) (∵ 1−sin2θ=cos2θ)
⇒ y=sin−1(2sinθ cosθ)=sin−1(sin 2θ)
⇒ y=2θ⇒y=2sin−1x
Differentiating both sides w.r.t. x, we get
⇒ dydx=2√1−x2 (∵ sin−1x=1√1−x2)