The derivative of
tan ⁻¹( (√(1 + x²) - 1)/x ) with
respect to tan ⁻¹( 2x√(1 - x²)/(1 - 2x²) ) at x = 1/2 is :

Solution

Maths sir · 4 steps

Put x = tan θ in one and x = sin φ in the other, and both inverse tans become plain angles.

1. Simplify u. Let u = tan⁻¹((√(1+x²)-1)/x) and put x = tan θ. The inside becomes (sec θ - 1)/(tan θ) = (1-cos θ)/(sin θ) = tan θ/2, so u = 1/2 tan⁻¹ x.
2. Simplify v. Let v = tan⁻¹(2x√(1-x²)/(1-2x²)) and put x = sin φ. The inside becomes (2 sin φ cos φ)/(cos 2φ) = tan 2φ. Near x = 1/2 we have φ = π/6 and 2φ = π/3, which lies in (-π/2, π/2), so v = 2 sin⁻¹ x.

   du/dv = (du/dx)/(dv/dx) = (1/2(1+x²))/(2/√(1-x²))

3. Put in numbers. At x = 1/2: du/dx = 1/(2 × 5/4) = 2/5 and dv/dx = 2/(√3/2) = 4/√3.
4. So. du/dv = 2/5 × √3/4 = 2√3/20 = √3/10, which is option (c).

The idea to keep. Before differentiating an ugly inverse-trig expression, look for the substitution that turns it into a plain angle. For the derivative of one function with respect to another, divide their x-derivatives.

Answer from the NTA final answer key. Solution by PrepWiser.