Q.Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$

Step 1: Let $x = \tan\theta$. Since $0 < x < 1$, $\theta \in \left(0, \tfrac{\pi}{4}\right)$. Then

$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta, \qquad y = \cos^{-1}(\cos 2\theta).$ …

Recognising the Identity

The expression $\dfrac{1-x^2}{1+x^2}$ is the double-angle formula $\cos 2\theta = \dfrac{1-\tan^2\theta}{1+\tan^2\theta}$ written in terms of $x = \tan\theta$. Substituting collapses the inverse cosine's argument to $\cos 2\theta$, but simplifying $\cos^{-1}(\cos 2\theta)$ to $2\theta$ requires $2\theta$ to sit inside $\cos^{-1}$'s principal range — which the given domain on $x$ guarantees.

Step-by-Step Solution

1. Substitute $x = \tan\theta$.

Since $0 < x < 1$, let $\theta = \tan^{-1}x \in \left(0, \tfrac{\pi}{4}\right)$ (because $\tan 0 = 0$ and $\tan\tfrac{\pi}{4} = 1$). Then:

2. Check the range of $2\theta$.

Since $\theta \in \left(0, \tfrac{\pi}{4}\right)$:

$2\theta \in \left(0, \frac{\pi}{2}\right) \subset [0, \pi],$ …

Method: Simplifying a Composite Inverse-Cosine Expression via Substitution

This is the double-angle substitution technique applied to $\cos^{-1}$ — the key difference from the $\sin^{-1}$ version is that $\cos^{-1}$'s principal range is $[0,\pi]$, not $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so the branch-checking step uses a different interval.

Steps

Step 1: Recognise the double-angle-cosine shape

$\frac{1-x^2}{1+x^2}$ matches $\cos 2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ exactly.

Step 2: Substitute $x=\tan\theta$

The expression becomes $y=\cos^{-1}(\cos 2\theta)$, where $\theta=\tan^{-1}x$.

Step 3: Work out the range of $2\theta$ from the given domain of $x$

Convert the stated restriction on $x$ into a restriction on $\theta=\tan^{-1}x$ (using that $\tan^{-1}$ is increasing), then double it to get the range of $2\theta$.

Step 4: Check that range against $\cos^{-1}$'s principal range $[0,\pi]$ — not $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ …

Common Mistakes

Mistake 1: Checking the wrong principal range

Students who have just practised $\sin^{-1}$ problems often reflexively check whether the angle lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, forgetting that $\cos^{-1}$'s principal range is different: $[0,\pi]$. Why it's wrong: using sine's range on a cosine problem can wrongly reject (or wrongly accept) a branch, leading to a sign or $\pi$-shift error. Correct approach: always identify which inverse function is on the outside first, and use its OWN principal range for the check.

Mistake 2: Forgetting to finish by differentiating the simplified $y$ …