Q.Find $\frac{dy}{dx}$, if $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}$, $0 < x < 1$

Concept: Second Derivative Inverse Cosine — the derivative of $\sin^{-1} \sqrt{1-x^2}$ simplifies using the identity $\sin^{-1} \sqrt{1-x^2} = \cos^{-1} x$ for $0 < x < 1$.

Step 1: For $0 < x < 1$, note that $\sqrt{1-x^2}$ is positive and in $(0,1)$. The principal value of $\sin^{-1} \sqrt{1-x^2}$ equals $\cos^{-1} x$, because $\sin(\cos^{-1} x) = \sqrt{1-x^2}$.

Step 2: Hence $y = \sin^{-1} x + \cos^{-1} x$. …

Why This Works: The Second Derivative Inverse Cosine Insight

When you see a sum of inverse trigonometric functions, your first instinct should be to check if they combine into a constant. For $0 < x < 1$, both $\sin^{-1} x$ and $\cos^{-1} x$ are defined and their sum is famously $\frac{\pi}{2}$. The trick here is recognizing that $\sin^{-1} \sqrt{1-x^2}$ is actually $\cos^{-1} x$ in disguise — but only for the given domain.

The domain $0 < x < 1$ is crucial. Outside this interval, the simplification changes sign or becomes undefined. Inside it, $\sqrt{1-x^2}$ is positive and less than 1, so the inverse sine is well-defined and yields an angle in $(0, \frac{\pi}{2})$.

Let's work through it step by step.

1. Set up the function

   We have $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}$, with $0 < x < 1$.

2. Simplify the second term

   Let $\theta = \sin^{-1} \sqrt{1-x^2}$. Then $\sin \theta = \sqrt{1-x^2}$.

   Since $0 < x < 1$, we have $0 < \sqrt{1-x^2} < 1$, so $\theta$ lies in $(0, \frac{\pi}{2})$.

   Now, $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$.

   Because $x > 0$, $|x| = x$, so $\cos \theta = x$.

   Since $\theta \in (0, \frac{\pi}{2})$, we have $\theta = \cos^{-1} x$.

   Watch out

   A common mistake is to forget the absolute value. If $x$ were negative, $\sqrt{x^2} = |x| = -x$, and the simplification would give $\theta = \cos^{-1}(-x) = \pi - \cos^{-1} x$, which changes the sum entirely. Always check the domain.

3. Rewrite the function

   Substituting back:

   $y = \sin^{-1} x + \cos^{-1} x$ …

Method: Spotting a Constant Before Differentiating

Before diving into the chain rule on a sum of inverse trig functions, check whether the expression itself simplifies to a constant using a standard identity — if it does, the derivative is $0$ immediately, with no calculus needed at all.

Steps

Step 1: Look for a recognisable inverse-trig identity in the given expression

Common ones: $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$, $\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$, and conversions like $\sin^{-1}\sqrt{1-x^2}=\cos^{-1}x$ (valid on a restricted domain).

Step 2: Justify the identity carefully using the given domain …

Common Mistakes

Mistake 1: Differentiating $\sin^{-1}\sqrt{1-x^2}$ directly via the chain rule instead of first checking for a constant.

Why it's wrong: the direct route is far messier (involving a nested square root inside an inverse sine) and much more likely to produce an algebra error than simply recognising the sum is constant. Correct approach: always scan a sum of inverse trig terms for a known identity before reaching for the chain rule.

Mistake 2: Forgetting the absolute value / domain check when converting $\sqrt{1-x^2}$-based inverse sine to inverse cosine. …