Match List-I with List-II (x > 0 wherever needed)
| List-I | List-II | ||
|---|---|---|---|
| Function | Derivative with respect to x | ||
| (A) | \(e^{2x}\) | (I) | \(2^{x}\log_e 2\) |
| (B) | \(\log_e(2x)\) | (II) | \(\displaystyle\frac{2}{1 + 4x^{2}}\) |
| (C) | \(2^{x}\) | (III) | \(\displaystyle\frac{1}{x}\) |
| (D) | \(\tan^{-1}(2x)\) | (IV) | \(2e^{2x}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \(e^{2x}\)
This matches (IV) \(2e^{2x}\). By the chain rule:
\( \displaystyle \frac{d}{dx}e^{2x} = e^{2x} \cdot 2 = 2e^{2x} \)
(B) \(\log_e(2x)\)
This matches (III) \(\frac{1}{x}\):
\( \displaystyle \frac{d}{dx}\log_e(2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \)
(C) \(2^{x}\)
This matches (I) \(2^{x}\log_e 2\), since \(\frac{d}{dx}a^{x} = a^{x}\log_e a\).
(D) \(\tan^{-1}(2x)\)
This matches (II) \(\frac{2}{1 + 4x^{2}}\):
\( \displaystyle \frac{d}{dx}\tan^{-1}(2x) = \frac{1}{1 + (2x)^{2}} \cdot 2 = \frac{2}{1 + 4x^{2}} \)