Match List-I with List-II

List-I (Curve)	List-II (Slope of tangent at $x = 4$)
(A) $y = \sqrt{x^3}$	(I) -1
(B) $y = \sqrt{x}$	(II) 1
(C) $y = x^3 - 47x$	(III) 1/4
(D) $xy = 16$	(IV) 3

Choose the correct answer from the options given below:

For $y = \sqrt{x^3}$:

$y = x^{3/2}$

$\frac{dy}{dx} = \frac{3}{2}x^{1/2}$

$\frac{dy}{dx} = \frac{3}{2}\sqrt{x}$

At $x = 4$:

$\frac{dy}{dx}\bigg|_{x=4} = \frac{3}{2}\sqrt{4}$

$= \frac{3}{2} \times 2$

$= 3$

(A) matches with (IV)

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For $y = \sqrt{x}$:

$y = x^{1/2}$

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

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

At $x = 4$:

$\frac{dy}{dx}\bigg|_{x=4} = \frac{1}{2\sqrt{4}}$

$= \frac{1}{2 \times 2}$

$= \frac{1}{4}$

(B) matches with (III)

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For $y = x^3 - 47x$:

$\frac{dy}{dx} = 3x^2 - 47$

At $x = 4$:

$\frac{dy}{dx}\bigg|_{x=4} = 3(4)^2 - 47$

$= 3 \times 16 - 47$

$= 48 - 47$

$= 1$

(C) matches with (II)

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For $xy = 16$:

$y = \frac{16}{x}$

$y = 16x^{-1}$

$\frac{dy}{dx} = -16x^{-2}$

$\frac{dy}{dx} = -\frac{16}{x^2}$

At $x = 4$:

$\frac{dy}{dx}\bigg|_{x=4} = -\frac{16}{4^2}$

$= -\frac{16}{16}$

$= -1$

(D) matches with (I)

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The correct matching is:

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)