If \(y = \sin^{-1}\left(\displaystyle\frac{1}{\sqrt{x + 1}}\right)\), \(x \gt 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\displaystyle\frac{1}{2\sqrt{x}(x + 1)}\)
Let \(u = (x + 1)^{-1/2}\). Then
\( \displaystyle \frac{du}{dx} = -\frac{1}{2}(x + 1)^{-3/2} \)
\( \displaystyle 1 - u^{2} = 1 - \frac{1}{x + 1} = \frac{x}{x + 1} \)
By the chain rule with \(\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^{2}}}\):
\( \displaystyle \frac{dy}{dx} = \frac{\sqrt{x + 1}}{\sqrt{x}} \cdot \left(-\frac{1}{2(x + 1)^{3/2}}\right) \)
\( \displaystyle = -\frac{1}{2\sqrt{x}(x + 1)} \)