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

Step 1: Let $x = \tan\theta$, so $\theta = \tan^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Then

Step 2: Matching $2\theta$ to the principal range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ of $\sin^{-1}$ gives the piecewise simplification:

Step 3: Differentiate each branch using $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$:

• For $|x| < 1$: $\dfrac{dy}{dx} = \dfrac{d}{dx}\big[2\tan^{-1}x\big] = \dfrac{2}{1+x^2}$. …

Why the Substitution Helps

The argument $\dfrac{2x}{1+x^2}$ is exactly the double-angle formula $\sin 2\theta = \dfrac{2\tan\theta}{1+\tan^2\theta}$ in disguise. Setting $x = \tan\theta$ turns the messy rational expression inside $\sin^{-1}$ into a clean $\sin 2\theta$, but simplifying $\sin^{-1}(\sin 2\theta)$ to $2\theta$ is only valid when $2\theta$ lies in the principal range of $\sin^{-1}$ — otherwise a correction of $\pm\pi$ is required. Getting this piecewise form right before differentiating is essential, since the derivative differs by sign across the pieces.

Step-by-Step Solution

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

Let $\theta = \tan^{-1}x$, so $\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Then:

2. Determine when $\sin^{-1}(\sin 2\theta) = 2\theta$ directly.

This holds only when $2\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, i.e. $\theta \in \left[-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right]$, i.e. $x = \tan\theta \in [-1, 1]$.

• Case $|x| \le 1$: $2\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, so $y = 2\theta = 2\tan^{-1}x$.
• Case $x > 1$: $\theta \in \left(\tfrac{\pi}{4}, \tfrac{\pi}{2}\right)$, so $2\theta \in \left(\tfrac{\pi}{2}, \pi\right)$. Using $\sin(\pi - 2\theta) = \sin 2\theta$ and $\pi - 2\theta \in \left(0, \tfrac{\pi}{2}\right)$ (inside the principal range), $y = \pi - 2\theta = \pi - 2\tan^{-1}x$.
• Case $x < -1$: $\theta \in \left(-\tfrac{\pi}{2}, -\tfrac{\pi}{4}\right)$, so $2\theta \in \left(-\pi, -\tfrac{\pi}{2}\right)$. Using $\sin(-\pi - 2\theta) = \sin 2\theta$ and $-\pi - 2\theta \in \left(-\tfrac{\pi}{2}, 0\right)$ (inside the principal range), $y = -\pi - 2\theta = -\pi - 2\tan^{-1}x$.

So:

3. Differentiate each branch. …

Method: Simplifying a Composite Inverse-Sine Expression by Trigonometric Substitution

This method applies whenever the expression inside an inverse trig function has the exact algebraic shape of a double-angle (or triple-angle) trig identity — substituting $x=\tan\theta$ turns the messy algebraic fraction into a single angle, which the outer inverse function can then simplify, PROVIDED that angle lies in the right principal range.

Steps

Step 1: Recognise the identity hiding in the expression

The fraction $\frac{2x}{1+x^2}$ has the exact form of $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}$. Whenever you see "$2x$ over $1+x^2$," "$1-x^2$ over $1+x^2$," or similar patterns, suspect a double-angle substitution.

Step 2: Substitute $x=\tan\theta$, i.e. $\theta=\tan^{-1}x$

This converts the algebraic fraction into a clean trig expression: $\frac{2x}{1+x^2}=\sin 2\theta$, so the original expression becomes $y=\sin^{-1}(\sin 2\theta)$.

Step 3: Find the range of $2\theta$ from the range of $\theta=\tan^{-1}x$

Since $\tan^{-1}x$ has range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for every real $x$, $2\theta$ ranges over $(-\pi,\pi)$ — wider than $\sin^{-1}$'s principal range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. You must therefore split into cases based on where $x$ places $2\theta$.

Step 4: Apply $\sin^{-1}(\sin\alpha)=\alpha$ only where valid, and correct it elsewhere …

Common Mistakes

Mistake 1: Writing $y=2\tan^{-1}x$ for every $x$, without checking the range

The temptation after substituting $x=\tan\theta$ is to declare $y=\sin^{-1}(\sin 2\theta)=2\theta$ unconditionally. Why it's wrong: $\sin^{-1}(\sin\alpha)=\alpha$ holds only when $\alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$; for $|x|>1$, $2\theta$ falls outside that range and the simplification needs a $\pi$-correction. Correct approach: always find the range of the doubled angle from the given $x$-restriction before cancelling the inverse and the trig function.

Mistake 2: Stopping at the simplified form of $y$ and treating it as the final answer …