Let $y=co{s}^{-1}!\left(\frac{1-x^2}{1+x^2}\right)$. Find $\frac{dy}{dx}$ and evaluate it at $x=\frac{1}{\sqrt{3}}$. Give the final numerical value.

Step-by-step solution

1. Write $u(x)=\frac{1-x^2}{1+x^2}$ so that $y=co{s}^{-1}u$.
2. Differentiate: $\frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}},\frac{du}{dx}$.
3. Compute $\frac{du}{dx}=\frac{(-2x)(1+x^2)-(1-x^2)(2x)}{(1+x^2{)}^2}=\frac{-4x}{(1+x^2{)}^2}$.
4. Compute $1-u^2$: $u^2=\frac{(1-x^2{)}^2}{(1+x^2{)}^2}$, so $1-u^2=\frac{(1+x^2{)}^2-(1-x^2{)}^2}{(1+x^2{)}^2}=\frac{4x^2}{(1+x^2{)}^2}$.
5. Thus $\sqrt{1-u^2}=\frac{2|x|}{1+x^2}$. For $x=\frac{1}{\sqrt{3}}>0$, $|x|=x$. Substitute: $\frac{dy}{dx}=-\frac{1}{\frac{2x}{1+x^2}}\cdot \frac{-4x}{(1+x^2{)}^2}=\frac{2}{1+x^2}$. At $x=\frac{1}{\sqrt{3}}$, $1+x^2=1+\frac{1}{3}=\frac{4}{3}$. Hence $\frac{dy}{dx}=\frac{2}{4/3}=\frac{3}{2}$.
6. Thus the value is $3/2$.

Concept tested: Derivative of arccos with rational argument

Published 18 September 2026 · How this solution was prepared and checked