If y = tan-1((3x − x3)/(1 − 3x2)), where −1/√3 < x < 1/√3, then dy/dx is

Concept

For any angle θ the tangent triple-angle identity gives tan 3θ = (3 tan θ − tan3θ)/(1 − 3 tan2θ). The inverse step tan-1(tan φ) = φ is legitimate only when φ lies inside the principal range (−π/2, π/2). The derivative fact needed at the end is d/dx tan-1x = 1/(1 + x2).

Applying the identity here

1. Substitute x = tan θ, so that θ = tan-1x. The given interval −1/√3 < x < 1/√3 corresponds to −π/6 < θ < π/6.

2. The argument then reads (3 tan θ − tan3θ)/(1 − 3 tan2θ), which by the identity is exactly tan 3θ.

3. From −π/6 < θ < π/6 it follows that −π/2 < 3θ < π/2, so 3θ already lies in the principal range and y = tan-1(tan 3θ) = 3θ; no ±π correction is needed.

4. Therefore y = 3θ = 3 tan-1x throughout the given interval.

5. Differentiate: dy/dx = 3 · d/dx(tan-1x) = 3/(1 + x2).

Cross-check by direct differentiation

The same value follows without any substitution. Write u = (3x − x3)/(1 − 3x2), so that y = tan-1u and dy/dx = u′/(1 + u2).

1. Quotient rule: u′ = [(3 − 3x2)(1 − 3x2) + 6x(3x − x3)]/(1 − 3x2)2 = 3(1 + x2)2/(1 − 3x2)2.

2. Denominator: 1 + u2 = [(1 − 3x2)2 + (3x − x3)2]/(1 − 3x2)2 = (1 + x2)3/(1 − 3x2)2.

3. Dividing the two, the (1 − 3x2)2 factors cancel and dy/dx = 3(1 + x2)2/(1 + x2)3 = 3/(1 + x2).

So dy/dx = 3/(1 + x2) on −1/√3 < x < 1/√3. A numerical check at x = 0 agrees: the derivative there is 3.