Question

Differentiate $y = \cos^{-1}\left( \frac{1 - x^2}{1 + x^2} \right)$ with respect to $x$, when $x \in (0, 1)$.

Hint:

When differentiating a function composed of a rational function inside an inverse trigonometric function, first simplify the expression for $u$, then apply the chain rule and quotient rule.

Answer 1

We are given: \[ y = \cos^{-1}\left( \frac{1 - x^2}{1 + x^2} \right). \] Let $u = \frac{1 - x^2}{1 + x^2}$. Using the chain rule, the derivative of $\cos^{-1} u$ is: \[ \frac{d}{dx} \cos^{-1} u = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \] Now, compute $\frac{du}{dx}$: \[ u = \frac{1 - x^2}{1 + x^2}, \] and apply the quotient rule to find: \[ \frac{du}{dx} = \frac{(2x)(1 + x^2) - (1 - x^2)(2x)}{(1 + x^2)^2}. \] Simplifying the numerator: \[ \frac{du}{dx} = \frac{2x(1 + x^2) - 2x(1 - x^2)}{(1 + x^2)^2} = \frac{4x^3}{(1 + x^2)^2}. \] Now, substitute this into the derivative formula for $\cos^{-1} u$: \[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - \left(\frac{1 - x^2}{1 + x^2}\right)^2}} \cdot \frac{4x^3}{(1 + x^2)^2}. \]