Find $\frac{dy}{dx}$ for $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ at $x=\frac{1}{\sqrt{3}}$. Give the numerical value.

Step-by-step solution

1. Let $u=\frac{1-x^2}{1+x^2}$. Then $y={\cos }^{-1}u$ and $\frac{dy}{dx}=\frac{-1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$ (derivative of ${\cos }^{-1}u$).
2. Compute $\frac{du}{dx}$ using quotient rule: $\frac{du}{dx}=\frac{-(2x)(1+x^2)-(1-x^2)(2x)}{(1+x^2{)}^2}=\frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2{)}^2}=\frac{-4x}{(1+x^2{)}^2}$.
3. Now find $u$ at $x=\frac{1}{\sqrt{3}}$: $x^2=\frac{1}{3}$, so $u=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{1}{2}$.
4. Compute $1-u^2=1-{\left(\frac{1}{2}\right)}^2=1-\frac{1}{4}=\frac{3}{4}$, thus $\sqrt{1-u^2}=\frac{\sqrt{3}}{2}$.
5. Evaluate $\frac{du}{dx}$ at the same $x$: $\frac{du}{dx}=\frac{-4x}{(1+x^2{)}^2}=\frac{-4\cdot \frac{1}{\sqrt{3}}}{{\left(1+\frac{1}{3}\right)}^2}=\frac{-\frac{4}{\sqrt{3}}}{{\left(\frac{4}{3}\right)}^2}=\frac{-\frac{4}{\sqrt{3}}}{\frac{16}{9}}=-\frac{4}{\sqrt{3}}\cdot \frac{9}{16}=-\frac{9}{4\sqrt{3}}$.
6. Now substitute into the derivative formula: $\frac{dy}{dx}=\frac{-1}{\frac{\sqrt{3}}{2}}\times \left(-\frac{9}{4\sqrt{3}}\right)=\left(-\frac{2}{\sqrt{3}}\right)\times \left(-\frac{9}{4\sqrt{3}}\right)=\frac{18}{4\cdot 3}=\frac{18}{12}=\frac{3}{2}$.
7. But notice a sign error: the formula $\frac{dy}{dx}=-\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$ already contains a negative; we used it, so the two negatives cancel, giving a positive result $\frac{3}{2}$. However the standard derivative of ${\cos }^{-1}$ is $-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$, and $\frac{du}{dx}$ is negative, so the final sign is negative: $\frac{dy}{dx}=-\frac{1}{\frac{\sqrt{3}}{2}}\times \left(-\frac{9}{4\sqrt{3}}\right)=-\left(-\frac{2}{\sqrt{3}}\right)\times \frac{9}{4\sqrt{3}}=-\left(\frac{-18}{12}\right)=-\left(-\frac{3}{2}\right)=\frac{3}{2}$. Correcting the arithmetic, the product of two negatives yields a positive, so the derivative at the point is $\frac{3}{2}$.
8. Since the question asks for the value, the simplified exact result is $\frac{3}{2}$.

Concept tested: Differentiation of inverse cosine using chain rule

Published 10 September 2026 · How this solution was prepared and checked