Match List-I with List-II:
| List-I | List-II | ||
| Function | Derivative w.r.t. x | ||
| (A) | \(\displaystyle\frac{5^x}{\log_e 5}\) | (I) | \(5^x(\log_e 5)^2\) |
| (B) | \(\log_e 5\) | (II) | \(5^x\log_e 5\) |
| (C) | \(5^x\log_e 5\) | (III) | \(5^x\) |
| (D) | \(5^x\) | (IV) | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
The basic result is \(\frac{d}{dx}\left(a^x\right) = a^x\log_e a\). Note that \(\log_e 5\) is a constant, so it simply stays as a factor.
(A) \(\frac{5^x}{\log_e 5}\)
This matches (III) \(5^x\).
\( \displaystyle \frac{d}{dx}\left(\frac{5^x}{\log_e 5}\right) = \frac{5^x\log_e 5}{\log_e 5} = 5^x \)
(B) \(\log_e 5\)
This matches (IV) 0, since the derivative of a constant is 0.
(C) \(5^x\log_e 5\)
This matches (I) \(5^x(\log_e 5)^2\).
\( \displaystyle \frac{d}{dx}\left(5^x\log_e 5\right) = \log_e 5 \times 5^x\log_e 5 = 5^x(\log_e 5)^2 \)
(D) \(5^x\)
This matches (II) \(5^x\log_e 5\).
\( \displaystyle \frac{d}{dx}\left(5^x\right) = 5^x\log_e 5 \)
So the matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).