If $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}, x \in (-1, 0)$, then $\frac{dy}{dx}$ is equal to

Given $y = \sin^{-1} x + \sin^{-1} \sqrt{1-x^2}$ where $x \in (-1, 0)$.

The derivative formula for inverse sine is:

$\frac{d}{dx}[\sin^{-1} u] = \frac{1}{\sqrt{1-u^2}} \times \frac{du}{dx}$

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For the first term:

$\frac{d}{dx}[\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}}$

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For the second term $\sin^{-1} \sqrt{1-x^2}$, let $u = \sqrt{1-x^2}$.

$u = (1-x^2)^{1/2}$

$\frac{du}{dx} = \frac{1}{2}(1-x^2)^{-1/2} \times (-2x)$

$\frac{du}{dx} = \frac{-x}{\sqrt{1-x^2}}$

Applying the chain rule:

$\frac{d}{dx}[\sin^{-1} \sqrt{1-x^2}] = \frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}} \times \frac{-x}{\sqrt{1-x^2}}$

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Simplifying the denominator:

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

$= \sqrt{x^2}$

$= |x|$

Since $x \in (-1, 0)$, we have $|x| = -x$.

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Continuing the simplification:

$\frac{d}{dx}[\sin^{-1} \sqrt{1-x^2}] = \frac{1}{-x} \times \frac{-x}{\sqrt{1-x^2}}$

$= \frac{-x}{-x \cdot \sqrt{1-x^2}}$

$= \frac{1}{\sqrt{1-x^2}}$

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Combining both derivatives:

$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$

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This is what NTA has marked right in the final answer key.