If \(y = \sin^{-1}\sqrt{\displaystyle\frac{x}{x+1}} + \sec^{-1}\sqrt{\displaystyle\frac{x+1}{x}}\), then \(\displaystyle\frac{dy}{dx}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
For \(|t| \ge 1\), \(\sec^{-1}t = \cos^{-1}\frac{1}{t}\). So the second term is
\( \displaystyle \sec^{-1}\sqrt{\frac{x+1}{x}} = \cos^{-1}\sqrt{\frac{x}{x+1}} \)
Let \(u = \sqrt{\frac{x}{x+1}}\), which lies between 0 and 1. Using the identity \(\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}\):
\( \displaystyle y = \sin^{-1}u + \cos^{-1}u = \frac{\pi}{2} \)
y is a constant, so \(\frac{dy}{dx} = 0\).