Match List-I with List-II (in (A), \(-1 \lt x \lt 1\); in (C), \(x \gt 1\))
| List-I | List-II | ||
|---|---|---|---|
| Function | Derivative with respect to x | ||
| (A) | \(\sin^{-1}(x^{2})\) | (I) | \(\displaystyle\frac{e^{x}}{1 + e^{2x}}\) |
| (B) | \(\tan^{-1}(e^{x})\) | (II) | \(-\displaystyle\frac{2x}{1 + x^{4}}\) |
| (C) | \(\cos^{-1}\left(\displaystyle\frac{1}{x}\right)\) | (III) | \(\displaystyle\frac{2x}{\sqrt{1 - x^{4}}}\) |
| (D) | \(\cot^{-1}(x^{2})\) | (IV) | \(\displaystyle\frac{1}{x\sqrt{x^{2} - 1}}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Each function is a standard inverse trigonometric derivative of an inner function u, multiplied by u′ (chain rule).
(A) \(\sin^{-1}(x^{2})\)
This matches (III) \(\frac{2x}{\sqrt{1 - x^{4}}}\). With \(u = x^{2}\):
\( \displaystyle \frac{1}{\sqrt{1 - u^{2}}} \cdot u' = \frac{2x}{\sqrt{1 - x^{4}}} \)
(B) \(\tan^{-1}(e^{x})\)
This matches (I) \(\frac{e^{x}}{1 + e^{2x}}\). With \(u = e^{x}\):
\( \displaystyle \frac{1}{1 + u^{2}} \cdot u' = \frac{e^{x}}{1 + e^{2x}} \)
(C) \(\cos^{-1}\left(\frac{1}{x}\right)\)
This matches (IV) \(\frac{1}{x\sqrt{x^{2} - 1}}\). With \(u = \frac{1}{x}\), \(u' = -\frac{1}{x^{2}}\):
\( \displaystyle -\frac{1}{\sqrt{1 - \frac{1}{x^{2}}}} \cdot \left(-\frac{1}{x^{2}}\right) = \frac{1}{x^{2}} \cdot \frac{x}{\sqrt{x^{2} - 1}} \)
\( \displaystyle = \frac{1}{x\sqrt{x^{2} - 1}} \)
(using \(\sqrt{x^{2}} = x\) for x > 1).
(D) \(\cot^{-1}(x^{2})\)
This matches (II) \(-\frac{2x}{1 + x^{4}}\). With \(u = x^{2}\):
\( \displaystyle -\frac{1}{1 + u^{2}} \cdot u' = -\frac{2x}{1 + x^{4}} \)