Match List-I with List-II
| List-I | List-II | ||
| (A) | \( \displaystyle\frac{d}{dx}\,2^{\sin x} \) | (I) | \( -\sin 2x \) |
| (B) | \( \displaystyle\frac{d}{dx}\,2^{\cos x} \) | (II) | \( 2^{\sin x}\log 2\ \cos x \) |
| (C) | \( \displaystyle\frac{d}{dx}\,e^{2\log\sin x} \) | (III) | \( 2^{\cos x}\log 2\ (-\sin x) \) |
| (D) | \( \displaystyle\frac{d}{dx}\,e^{2\log\cos x} \) | (IV) | \( \sin 2x \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
For (A) and (B) we use the rule \( \frac{d}{dx}a^{u}=a^{u}\log a\cdot\frac{du}{dx} \). For (C) and (D), since \( e^{\log f}=f \), the function simplifies before differentiating.
(A) \( \displaystyle\frac{d}{dx}\,2^{\sin x} \)
This matches (II).
\( \displaystyle \frac{d}{dx}\,2^{\sin x}=2^{\sin x}\log 2\ \cos x \)
(B) \( \displaystyle\frac{d}{dx}\,2^{\cos x} \)
This matches (III).
\( \displaystyle \frac{d}{dx}\,2^{\cos x}=2^{\cos x}\log 2\ (-\sin x) \)
(C) \( \displaystyle\frac{d}{dx}\,e^{2\log\sin x} \)
This matches (IV). Here \( e^{2\log\sin x}=\sin^{2}x \).
\( \displaystyle \frac{d}{dx}\sin^{2}x=2\sin x\cos x \)
\( \displaystyle =\sin 2x \)
(D) \( \displaystyle\frac{d}{dx}\,e^{2\log\cos x} \)
This matches (I). Here \( e^{2\log\cos x}=\cos^{2}x \).
\( \displaystyle \frac{d}{dx}\cos^{2}x=-2\sin x\cos x \)
\( \displaystyle =-\sin 2x \)
So the matching is (A) - (II), (B) - (III), (C) - (IV), (D) - (I).