Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \displaystyle\frac{d}{dx}\left(\sqrt{\sin x}\right) \) | (I) | \( \displaystyle\frac{-\sin x}{2\sqrt{\cos x}} \) |
| (B) | \( \displaystyle\frac{d}{dx}\left(\sqrt{\cos x}\right) \) | (II) | \( -\sin 2x \) |
| (C) | \( \displaystyle\frac{d}{dx}\left(\sin^{2}x\right) \) | (III) | \( \displaystyle\frac{\cos x}{2\sqrt{\sin x}} \) |
| (D) | \( \displaystyle\frac{d}{dx}\left(\cos^{2}x\right) \) | (IV) | \( \sin 2x \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Each is a chain rule derivative: the derivative of the outer function times the derivative of the inner function.
(A) \( \frac{d}{dx}\left(\sqrt{\sin x}\right) \)
This matches (III) \( \frac{\cos x}{2\sqrt{\sin x}} \).
\( \displaystyle \frac{1}{2\sqrt{\sin x}}\cdot\cos x=\frac{\cos x}{2\sqrt{\sin x}} \)
(B) \( \frac{d}{dx}\left(\sqrt{\cos x}\right) \)
This matches (I) \( \frac{-\sin x}{2\sqrt{\cos x}} \).
\( \displaystyle \frac{1}{2\sqrt{\cos x}}\cdot(-\sin x)=\frac{-\sin x}{2\sqrt{\cos x}} \)
(C) \( \frac{d}{dx}\left(\sin^{2}x\right) \)
This matches (IV) \( \sin 2x \).
\( \displaystyle 2\sin x\cos x=\sin 2x \)
(D) \( \frac{d}{dx}\left(\cos^{2}x\right) \)
This matches (II) \( -\sin 2x \).
\( \displaystyle 2\cos x\cdot(-\sin x)=-\sin 2x \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
Shortcut: Since \( \sin^2x+\cos^2x=1 \), the derivatives in (C) and (D) must add up to 0, so they are \( \sin 2x \) and \( -\sin 2x \); and \( \sin^2x \) increases near x = 0, so (C) - (IV). Only Option (3) has (C) - (IV) and (D) - (II).