Question

Match List-I with List-II.

List-I	List-II
A. \(\dfrac{d}{dx}(\cos^{-1}x)\)	I. \(\dfrac{1}{1+x^2}\)
B. \(\dfrac{d}{dx}(\cot^{-1}x)\)	II. \(\dfrac{1}{|x|\sqrt{x^2-1}}\)
C. \(\dfrac{d}{dx}(\cosec^{-1}x)\)	III. \(\dfrac{-1}{\sqrt{1-x^2}}\)
D. \(\dfrac{d}{dx}(\sec^{-1}x)\)	IV. \(\dfrac{-1}{|x|\sqrt{x^2-1}}\)

Choose the correct answer from the options given below:

• A. A-I, B-II, C-III, D-IV
• B. A-II, B-III, C-IV, D-I
• C. A-III, B-I, C-IV, D-II
• D. A-III, B-I, C-IV, D-II

Hint:

Memorize inverse trigonometric derivatives carefully—sign and modulus are very important.

Answer 1

The Correct Option is C

Concept: Standard derivatives of inverse trigonometric functions: \[ \frac{d}{dx}(\cos^{-1}x) = \frac{-1}{\sqrt{1-x^2}} \] \[ \frac{d}{dx}(\cot^{-1}x) = \frac{-1}{1+x^2} \] \[ \frac{d}{dx}(\cosec^{-1}x) = \frac{-1}{|x|\sqrt{x^2-1}} \] \[ \frac{d}{dx}(\sec^{-1}x) = \frac{1}{|x|\sqrt{x^2-1}} \]

Step 1: Match A. \[ \frac{d}{dx}(\cos^{-1}x) = \frac{-1}{\sqrt{1-x^2}} \Rightarrow A \rightarrow III \]

Step 2: Match B. \[ \frac{d}{dx}(\cot^{-1}x) = \frac{-1}{1+x^2} \Rightarrow B \rightarrow I \]

Step 3: Match C. \[ \frac{d}{dx}(\cosec^{-1}x) = \frac{-1}{|x|\sqrt{x^2-1}} \Rightarrow C \rightarrow IV \]

Step 4: Match D. \[ \frac{d}{dx}(\sec^{-1}x) = \frac{1}{|x|\sqrt{x^2-1}} \Rightarrow D \rightarrow II \]

Step 5: Final mapping. \[ A-III,\ B-I,\ C-IV,\ D-II \] \[ \boxed{(3)} \]