Let $y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$. Find $\frac{dy}{dx}$ and then evaluate it at $x=\frac{1}{2}$.

Step-by-step solution

1. Set $u=\frac{2x}{1-x^2}$, so $y={\tan }^{-1}u$.
2. Differentiate: $\frac{dy}{dx}=\frac{u^{\prime }}{1+u^2}$.
3. Compute $u^{\prime }$ using quotient rule: $u^{\prime }=\frac{2(1-x^2)-2x(-2x)}{(1-x^2{)}^2}=\frac{2(1-x^2+2x^2)}{(1-x^2{)}^2}=\frac{2(1+x^2)}{(1-x^2{)}^2}$.
4. Find $1+u^2=1+\frac{4x^2}{(1-x^2{)}^2}=\frac{(1-x^2{)}^2+4x^2}{(1-x^2{)}^2}=\frac{1+2x^2+x^4}{(1-x^2{)}^2}=\frac{(1+x^2{)}^2}{(1-x^2{)}^2}$.
5. Thus $\frac{dy}{dx}=\frac{\frac{2(1+x^2)}{(1-x^2{)}^2}}{\frac{(1+x^2{)}^2}{(1-x^2{)}^2}}=\frac{2}{1+x^2}$.
6. Now substitute $x=\frac{1}{2}$: $\frac{dy}{dx}=\frac{2}{1+{\left(\frac{1}{2}\right)}^2}=\frac{2}{1+\frac{1}{4}}=\frac{2}{\frac{5}{4}}=\frac{8}{5}=\frac{4}{5}$.

Concept tested: Derivative of composite inverse tangent

Published 16 September 2026 · How this solution was prepared and checked