The golden rule for inverse trig differentiation in JEE: SIMPLIFY THE EXPRESSION FIRST, then differentiate.

Key Identities (put x = tan t):

• tan^(-1)(2$\frac{x}{1-x^2}$) = 2*tan^(-1)(x) [for |x| < 1]
• sin^(-1)(2$\frac{x}{1+x^2}$) = 2*tan^(-1)(x) [for |x| <= 1]
• cos^(-1)$\frac{(1-x^2}{(1+x^2)}$) = 2*tan^(-1)(x) [for x >= 0]

Key Identities (put x = sin t or cos t):

• sin^(-1)(2xsqrt(1-$x^2$)) = 2sin^(-1)(x) [for |x| <= 1/sqrt(2)]
• sin^(-1)(3x - 4$x^3$) = 3*sin^(-1)(x) [for |x| <= 1/2]
• cos^(-1)(4$x^3$ - 3x) = 3*cos^(-1)(x) [for 1/2 <= x <= 1]

CRITICAL WARNING: The domain matters! sin^(-1)(sin(2t)) = 2t ONLY when -pi/2 <= 2t <= pi/2. Outside this range, sin^(-1)(sin(2t)) = pi - 2t or -(pi + 2t). Getting the domain wrong is the #1 error in these problems.

After simplification, derivatives become trivial: d/dx(2*tan^(-1)(x)) = $\frac{2}{1+x^2}$.