Question

If \( y = \sin^{-1}\left(\frac{1}{\sqrt{x+1}}\right) \) then \( \frac{dy}{dx} = \)

• A. \( \frac{1}{2\sqrt{1-x}} \)
• B. \( \frac{1}{2\sqrt{x}(1+\sqrt{x})} \)
• C. \( \frac{1}{2x(1+\sqrt{x})} \)
• D. \( -\frac{1}{2\sqrt{x}(1+x)} \)

Hint:

For inverse trigonometric derivatives, always simplify \( 1-u^2 \) carefully before substituting.

Answer 1

The Correct Option is D

Step 1: Let the inner function be \( u \).
\[ y = \sin^{-1}(u), \quad u = \frac{1}{\sqrt{x+1}}. \]

Step 2: Use derivative of inverse sine.
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}. \]

Step 3: Simplify \( u^2 \).
\[ u^2 = \frac{1}{x+1}. \]
\[ 1 - u^2 = \frac{x}{x+1}. \]
\[ \sqrt{1-u^2} = \sqrt{\frac{x}{x+1}} = \frac{\sqrt{x}}{\sqrt{x+1}}. \]

Step 4: Find \( \frac{du}{dx} \).
\[ u = (x+1)^{-1/2} \]
\[ \frac{du}{dx} = -\frac{1}{2}(x+1)^{-3/2}. \]

Step 5: Substitute into derivative.
\[ \frac{dy}{dx} = \frac{1}{\frac{\sqrt{x}}{\sqrt{x+1}}} \cdot \left(-\frac{1}{2}(x+1)^{-3/2}\right). \]

Step 6: Simplify expression.
\[ = \frac{\sqrt{x+1}}{\sqrt{x}} \cdot \left(-\frac{1}{2(x+1)^{3/2}}\right). \]
\[ = -\frac{1}{2\sqrt{x}(x+1)}. \]

Step 7: Final conclusion.
\[ \boxed{-\frac{1}{2\sqrt{x}(1+x)}} \]