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

Question

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

Accepted Answer

> [!TLDR] > The key idea is that for $0 < x < 1$, the second term $\sin^{-1} \sqrt{1-x^2}$ simplifies to $\cos^{-1} x$, and since $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, the function is constant. Therefore, $\frac{dy}{dx} = 0$. ## 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 $ heta = \sin^{-1} \sqrt{1-x^2}$. Then $\sin heta = \sqrt{1-x^2}$. Since $0 < x < 1$, we have $0 < \sqrt{1-x^2} < 1$, so $ heta$ lies in $(0, \frac{\pi}{2})$. Now, $\cos heta = \sqrt{1 - \sin^2 heta} = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$. Because $x > 0$, $|x| = x$, so $\cos heta = x$. Since $ heta \in (0, \frac{\pi}{2})$, we have $ heta = \cos^{-1} x$. > [!WARNING] > A common mistake is to forget the absolute value. If $x$ were negative, $\sqrt{x^2} = |x| = -x$, and the simplification would give $ heta = \cos^{-1}(-x) = \pi - \cos^{-1} x$, which changes the sum entirely. Always check the domain. 3. **Rewrite the function** Substituting back: $$y = \sin^{-1} x + \cos^{-1} x$$ …

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

Concept understanding — Inverse Function Relationship

Inverse Function Relationship

The defining equations

When does an inverse exist?

The geometry

Why the restriction bites — the trig case

Why This Works: The Second Derivative Inverse Cosine Insight

Method: Spotting a Constant Before Differentiating

Steps

Common Mistakes

Q.Find dydx\frac{dy}{dx}dxdy​, if y=sin⁡−1x+sin⁡−11−x2y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}y=sin−1x+sin−11−x2​, 0<x<10 < x < 10<x<1

Why This Works: The Second Derivative Inverse Cosine Insight

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

Unlock everything free for 14 days