Consider the following statements:
(A) \(\displaystyle\frac{d}{dx}\left(\sin^{-1}x + \cos^{-1}x\right) = 0\), \(-1 \lt x \lt 1\)
(B) \(\displaystyle\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{1 + x^{2}}\)
(C) \(\displaystyle\frac{d}{dx}\left(\cot^{-1}x\right) = \frac{1}{1 + x^{2}}\)
(D) \(\displaystyle\frac{d}{dx}\left(\sec^{-1}x\right) = \frac{1}{x\sqrt{x^{2} - 1}}\), \(x \gt 1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
The standard derivatives of the inverse trigonometric functions are
\( \displaystyle \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^{2}}}, \quad \frac{d}{dx}\cos^{-1}x = -\frac{1}{\sqrt{1 - x^{2}}} \)
\( \displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^{2}}, \quad \frac{d}{dx}\cot^{-1}x = -\frac{1}{1 + x^{2}} \)
\( \displaystyle \frac{d}{dx}\sec^{-1}x = \frac{1}{|x|\sqrt{x^{2} - 1}} \)
So the derivatives of \(\sin^{-1}x\) and \(\cos^{-1}x\) cancel, \(\cot^{-1}x\) has a minus sign, and for x > 1 the modulus in the \(\sec^{-1}x\) formula is just x.
Checking the statements
(A) \(\frac{d}{dx}\left(\sin^{-1}x + \cos^{-1}x\right) = 0\), \(-1 \lt x \lt 1\) — Correct. \(\frac{1}{\sqrt{1 - x^{2}}} - \frac{1}{\sqrt{1 - x^{2}}} = 0\).
(B) \(\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{1 + x^{2}}\) — Correct. This is the standard result.
(C) \(\frac{d}{dx}\left(\cot^{-1}x\right) = \frac{1}{1 + x^{2}}\) — Incorrect. The derivative is \(-\frac{1}{1 + x^{2}}\).
(D) \(\frac{d}{dx}\left(\sec^{-1}x\right) = \frac{1}{x\sqrt{x^{2} - 1}}\), \(x \gt 1\) — Correct. For x > 1, |x| = x.