Match List-I with List-II
| List-I | List-II | ||
| (Function) | (Derivative with respect to 'x') | ||
| (A) | \(f(x) = x^x\) | (I) | \(ax^{a-1}\) |
| (B) | \(f(x) = a^x\) | (II) | 0 |
| (C) | \(f(x) = a^a\) | (III) | \(a^x\log_e a\) |
| (D) | \(f(x) = x^a\) | (IV) | \(x^x(1+ \log_e x)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
The four functions differ in where the variable sits: in both base and exponent, only in the exponent, nowhere, or only in the base.
(A) \(f(x) = x^x\)
This matches (IV) \(x^x(1+ \log_e x)\). The variable is in both base and exponent, so logarithmic differentiation is used. Let \(y = x^x\):
\(\log_e y = x\log_e x\)
\( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log_e x + x \cdot \frac{1}{x} = 1 + \log_e x \)
\( \displaystyle \frac{dy}{dx} = x^x(1 + \log_e x) \)
(B) \(f(x) = a^x\)
This matches (III) \(a^x\log_e a\), the derivative of an exponential function with constant base a.
(C) \(f(x) = a^a\)
This matches (II) 0. \(a^a\) contains no x, so it is a constant and its derivative is 0.
(D) \(f(x) = x^a\)
This matches (I) \(ax^{a-1}\), by the power rule.
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).