Correct option is A. 2
Let u = cos–1 (2x2 – 1) and v = cos–1 x
\(\cfrac{du}{dv}=?\)
Considering u = cos–1 (2x2 – 1)
Put x = cos θ ⇒ θ = cos-1x ---(1)
u = cos–1 (2cos2θ – 1)
u = cos–1 (cos2θ)
∵ 2cos2θ – 1 = cos2θ
u = 2θ
u = 2 cos-1x -- From(1)
Differentiating w.r.t x we get,