Match List-I with List-II (in (A) and (B), \(-1 \lt x \lt 1\))
| List-I | List-II | ||
|---|---|---|---|
| Function | Derivative with respect to x | ||
| (A) | \(\log_e\left(\displaystyle\frac{1 - x^{2}}{1 + x^{2}}\right)\) | (I) | \(\displaystyle\frac{1}{1 - x^{2}}\) |
| (B) | \(\log_e\sqrt{\displaystyle\frac{1 + x}{1 - x}}\) | (II) | \(\displaystyle\frac{1}{1 + e^{x}}\) |
| (C) | \(\log_e\left(x + \sqrt{x^{2} + 1}\right)\) | (III) | \(-\displaystyle\frac{4x}{1 - x^{4}}\) |
| (D) | \(\log_e\left(\displaystyle\frac{e^{x}}{1 + e^{x}}\right)\) | (IV) | \(\displaystyle\frac{1}{\sqrt{x^{2} + 1}}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Splitting each logarithm with the laws of logarithms makes the differentiation short.
(A) \(\log_e\left(\frac{1 - x^{2}}{1 + x^{2}}\right)\)
This matches (III) \(-\frac{4x}{1 - x^{4}}\). Writing it as \(\log_e(1 - x^{2}) - \log_e(1 + x^{2})\):
\( \displaystyle \frac{dy}{dx} = \frac{-2x}{1 - x^{2}} - \frac{2x}{1 + x^{2}} \)
\( \displaystyle = \frac{-2x(1 + x^{2}) - 2x(1 - x^{2})}{1 - x^{4}} = -\frac{4x}{1 - x^{4}} \)
(B) \(\log_e\sqrt{\frac{1 + x}{1 - x}}\)
This matches (I) \(\frac{1}{1 - x^{2}}\). Writing it as \(\frac{1}{2}\left[\log_e(1 + x) - \log_e(1 - x)\right]\):
\( \displaystyle \frac{dy}{dx} = \frac{1}{2}\left[\frac{1}{1 + x} + \frac{1}{1 - x}\right] \)
\( \displaystyle = \frac{1}{2} \cdot \frac{2}{1 - x^{2}} = \frac{1}{1 - x^{2}} \)
(C) \(\log_e\left(x + \sqrt{x^{2} + 1}\right)\)
This matches (IV) \(\frac{1}{\sqrt{x^{2} + 1}}\). By the chain rule:
\( \displaystyle \frac{dy}{dx} = \frac{1}{x + \sqrt{x^{2} + 1}} \cdot \left(1 + \frac{x}{\sqrt{x^{2} + 1}}\right) \)
\( \displaystyle = \frac{1}{x + \sqrt{x^{2} + 1}} \cdot \frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}} = \frac{1}{\sqrt{x^{2} + 1}} \)
(D) \(\log_e\left(\frac{e^{x}}{1 + e^{x}}\right)\)
This matches (II) \(\frac{1}{1 + e^{x}}\). Writing it as \(x - \log_e(1 + e^{x})\):
\( \displaystyle \frac{dy}{dx} = 1 - \frac{e^{x}}{1 + e^{x}} = \frac{1}{1 + e^{x}} \)