For the function \( (ex)^{y}=e^{x} \);
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( y \) is equal to | I. | \( \displaystyle\frac{1-\log_{e}x}{x(1+\log_{e}x)^{3}} \) |
| B. | value of \( \displaystyle\frac{dy}{dx} \) | II. | \( \displaystyle\frac{x}{1+\log_{e}x} \) |
| C. | value of \( \displaystyle\frac{d^{2}y}{dx^{2}} \) | III. | \( \displaystyle\frac{\log_{e}x}{(1+\log_{e}x)^{2}} \) |
| D. | \( x^{y} \) is equal to | IV. | \( e^{x-y} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-III, C-I, D-IV
Taking logarithms of \( (ex)^{y}=e^{x} \),
\( \displaystyle y(1+\log_{e}x)=x \)
(A) So \( y=\frac{x}{1+\log_{e}x} \), which matches (II).
(B) By the quotient rule:
\( \displaystyle \frac{dy}{dx}=\frac{(1+\log_{e}x)-x\cdot\frac{1}{x}}{(1+\log_{e}x)^{2}} \)
\( \displaystyle =\frac{\log_{e}x}{(1+\log_{e}x)^{2}} \)
This matches (III).
(C) Differentiating \( \log_{e}x\,(1+\log_{e}x)^{-2} \) by the product rule:
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{1}{x}(1+\log_{e}x)^{-2}-\frac{2\log_{e}x}{x}(1+\log_{e}x)^{-3} \)
\( \displaystyle =\frac{(1+\log_{e}x)-2\log_{e}x}{x(1+\log_{e}x)^{3}} \)
\( \displaystyle =\frac{1-\log_{e}x}{x(1+\log_{e}x)^{3}} \)
This matches (I).
(D) From \( y+y\log_{e}x=x \), \( y\log_{e}x=x-y \), so
\( \displaystyle x^{y}=e^{y\log_{e}x} \)
\( \displaystyle =e^{x-y} \)
This matches (IV).
So the matching is (A)-(II), (B)-(III), (C)-(I), (D)-(IV).