Question:medium

Match List-I with List-II:
List-I (Function)List-II (Derivative w.r.t. x)
(A) \( \frac{5^x}{\ln 5} \)(I) \(5^x (\ln 5)^2\)
(B) \(\ln 5\)(II) \(5^x \ln 5\) 
(C) \(5^x \ln 5\)(III) \(5^x\) 
(D) \(5^x\)(IV) 0 

Choose the correct answer from the options given below.

Show Hint

When calculating derivatives of exponential functions like \( 5^x \), remember that the derivative of \( a^x \) is \( a^x \ln a \). For constants such as \( \log_e 5 \), their derivative is zero. Applying the chain rule or constant rule will help simplify the calculations.

Updated On: Jan 16, 2026
  • (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
  • (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
  • (A) - (I), (B) - (IV), (C) - (II), (D) - (III)
  • (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
Show Solution

The Correct Option is D

Solution and Explanation

To match the functions in List-I with their derivatives in List-II, the derivatives are calculated as follows:

For (A) \(f(x) = \frac{5^x}{\log_e 5}\):

\(f'(x) = \frac{d}{dx} \left(\frac{5^x}{\log_e 5}\right) = \frac{1}{\log_e 5} \frac{d}{dx}(5^x) = \frac{1}{\log_e 5} (5^x \log_e 5) = 5^x.\)

Therefore, (A) matches with (III).

For (B) \(f(x) = \log_e 5\):

\(f'(x) = \frac{d}{dx}(\log_e 5) = 0\) (since \(\log_e 5\) is a constant).

Therefore, (B) matches with (IV).

For (C) \(f(x) = 5^x \log_e 5\):

\(f'(x) = \frac{d}{dx}(5^x \log_e 5) = \log_e 5 \frac{d}{dx}(5^x) = \log_e 5 (5^x \log_e 5) = 5^x (\log_e 5)^2.\)

Therefore, (C) matches with (I).

For (D) \(f(x) = 5^x\):

\(f'(x) = \frac{d}{dx}(5^x) = 5^x \log_e 5.\)

Therefore, (D) matches with (II).

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

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