Match List-I with List-II

List-I	List-II
Function	Derivative
---	---
(A) $y = \sin^{-1} x + \sin^{-1} \sqrt{1 - x^2}; \vert x\vert < 1$	(I) $\frac{dy}{dx} = \frac{1}{2y-1}$
(B) $y = \sqrt{x + y}, x+y > 0 \text{ and } y \neq \frac{1}{2}$	(II) $\frac{dy}{dx} = 10^x \log_e 10$
(C) $y = \log_{10} x, x > 0$	(III) $\frac{dy}{dx} = 0$
(D) $y = 10^x$	(IV) $\frac{dy}{dx} = \frac{1}{x \log_e 10}$

Choose the correct answer from the options given below:

(A) $y = \sin^{-1}x + \sin^{-1}\sqrt{1 - x^2}$

Let $x = \sin\theta$, where $\theta \in \left[0, \frac{\pi}{2}\right)$

$\sqrt{1 - x^2} = \sqrt{1 - \sin^2\theta} = \cos\theta$

$y = \sin^{-1}(\sin\theta) + \sin^{-1}(\cos\theta)$

$y = \theta + \sin^{-1}\left(\sin\left(\frac{\pi}{2} - \theta\right)\right)$

$y = \theta + \frac{\pi}{2} - \theta$

$y = \frac{\pi}{2}$ (a constant)

$\frac{dy}{dx} = 0$

(A) → (III)

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(B) $y = \sqrt{x + y}$

$y^2 = x + y$

Differentiating both sides with respect to $x$:

$2y \cdot \frac{dy}{dx} = 1 + \frac{dy}{dx}$

$2y\frac{dy}{dx} - \frac{dy}{dx} = 1$

$(2y - 1)\frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{2y - 1}$

Since $y \neq \frac{1}{2}$, division by $(2y-1)$ is valid.

(B) → (I)

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(C) $y = \log_{10}x$

Using $\frac{d}{dx}(\log_a x) = \frac{1}{x \cdot \log_e a}$ with $a = 10$:

$\frac{dy}{dx} = \frac{1}{x \cdot \log_e 10}$

(C) → (IV)

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(D) $y = 10^x$

Using $\frac{d}{dx}(a^x) = a^x \cdot \log_e a$ with $a = 10$:

$\frac{dy}{dx} = 10^x \cdot \log_e 10$

(D) → (II)

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(A) → (III) $\frac{dy}{dx} = 0$

(B) → (I) $\frac{dy}{dx} = \frac{1}{2y-1}$

(C) → (IV) $\frac{dy}{dx} = \frac{1}{x\log_e 10}$

(D) → (II) $\frac{dy}{dx} = 10^x \log_e 10$

Therefore, the correct answer is Option 1.