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

Step 1: Let $x = \sin\theta$. Since $-\tfrac{1}{\sqrt2} < x < \tfrac{1}{\sqrt2}$, $\theta \in \left(-\tfrac{\pi}{4},\tfrac{\pi}{4}\right)$, so $\cos\theta > 0$ and $\sqrt{1-x^2} = \cos\theta$. Then

$2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta, \qquad y = \sin^{-1}(\sin 2\theta).$ …

Recognising the Identity

The expression $2x\sqrt{1-x^2}$ is the double-angle formula $\sin 2\theta = 2\sin\theta\cos\theta$ in terms of $x = \sin\theta$: once $\sqrt{1-x^2}$ is identified as $\cos\theta$, the whole argument collapses to $\sin 2\theta$.

Step-by-Step Solution

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

Let $\theta = \sin^{-1}x \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Since $-\tfrac{1}{\sqrt2} < x < \tfrac{1}{\sqrt2}$, we get $\theta \in \left(-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right)$, so $\cos\theta > 0$ throughout, and:

Hence:

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

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

exactly the principal range of $\sin^{-1}$. So $\sin^{-1}(\sin 2\theta) = 2\theta$ directly, with no correction needed.

3. Write $y$ in terms of $x$.

4. Differentiate.

Using $\dfrac{d}{dx}\sin^{-1}x = \dfrac{1}{\sqrt{1-x^2}}$: …

Method: Simplify Inverse-Trig Compositions by Trigonometric Substitution

This method applies whenever you must differentiate an inverse trig function whose argument is built from $x$ and $\sqrt{1-x^2}$ using a double-angle-style pattern — direct differentiation of the raw expression would be painful, so simplify first.

Steps

Step 1: Spot the double-angle pattern and substitute

Look at the argument inside the inverse trig function. An expression like $2x\sqrt{1-x^2}$ matches $2\sin\theta\cos\theta = \sin 2\theta$. Substitute $x = \sin\theta$, restricting $\theta$ to the principal branch of the inverse function you substituted with.

Step 2: Rewrite the argument using the identity

so the whole argument collapses to a single trig function of $2\theta$.

Step 3: Reduce using the principal-value range …

Common Mistakes

Mistake 1: Stopping at the simplified expression for $y$ instead of differentiating

The substitution trick reduces $y$ to something clean like $2\sin^{-1}x$ — but the question asks for $\dfrac{dy}{dx}$, not for a simplified $y$. Why it's wrong: simplifying is only the first half of the problem. Correct approach: always take the extra step and differentiate the simplified form, e.g. $\dfrac{d}{dx}(2\sin^{-1}x) = \dfrac{2}{\sqrt{1-x^2}}$.

Mistake 2: Ignoring the domain restriction when reducing $\sin^{-1}(\sin 2\theta)$ to $2\theta$ …