If y = sin^-1x + sin^-1√(1-x^2), x ∈ (-1,0) , then displaystyledy/dx…

The correct answer is Option (3) → \(\displaystyle\frac{2}{\sqrt{1-x^2}}\)

The first term gives

    \(\displaystyle \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}\)

For the second term let \(u = \sqrt{1 - x^2}\). Then

    \(\displaystyle \frac{du}{dx} = \frac{-x}{\sqrt{1-x^2}}\)

    \(\displaystyle \sqrt{1 - u^2} = \sqrt{1 - (1 - x^2)} = \sqrt{x^2} = |x|\)

Since \(x \in (-1, 0)\), x is negative, so \(|x| = -x\). By the chain rule,

    \(\displaystyle \frac{d}{dx}\sin^{-1}u = \frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx} = \frac{1}{-x}\cdot\frac{-x}{\sqrt{1-x^2}}\)

    \(\displaystyle = \frac{1}{\sqrt{1-x^2}}\)

Adding the two terms:

    \(\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}\)

Tip: For \(x \in (0, 1)\), \(|x| = x\), the second term has derivative \(-\frac{1}{\sqrt{1-x^2}}\) and dy/dx = 0. The interval decides the sign of \(\sqrt{x^2}\), so always check it.