Derivative of `sin^(-1) ((1)/(sqrt(x + 1)))` with respect to x is

Options

A. `(1)/(sqrt(x)(1 + x))`
B. `(-2)/(sqrt(x)(1 +x))`
C. ` (-1)/(2sqrt(x)(1 +x))`
D. None of these

Explanation

To find the derivative of \( y = \sin^{-1} \left( \frac{1}{\sqrt{x + 1}} \right) \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the Function
Let \( y = \sin^{-1} \left( \frac{1}{\sqrt{x + 1}} \right) \).

### Step 2: Differentiate using Chain Rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( \sin^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{1}{\sqrt{x + 1}} \).

### Step 3: Find \( \frac{du}{dx} \)
First, we need to differentiate \( u \):
\[ u = (x + 1)^{-1/2} \]
Using the power rule:
\[ \frac{du}{dx} = -\frac{1}{2} (x + 1)^{-3/2} \cdot (1) = -\frac{1}{2\sqrt{(x + 1)^3}} \]

### Step 4: Substitute \( u \) into the Derivative
Now we substitute \( u \) back into the derivative:
\[ 1 - u^2 = 1 - \left( \frac{1}{\sqrt{x + 1}} \right)^2 = 1 - \frac{1}{x + 1} = \frac{x + 1 - 1}{x + 1} = \frac{x}{x + 1} \]
Thus,
\[ \sqrt{1 - u^2} = \sqrt{\frac{x}{x + 1}} = \frac{\sqrt{x}}{\sqrt{x + 1}} \]

### Step 5: Combine the Derivatives
Now we can write the derivative:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} = \frac{1}{\frac{\sqrt{x}}{\sqrt{x + 1}}} \cdot \left(-\frac{1}{2\sqrt{(x + 1)^3}}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sqrt{x + 1}}{2\sqrt{x} \cdot (x + 1)^{3/2}} = -\frac{1}{2\sqrt{x} \cdot (x + 1)} \]

### Final Answer
Thus, the derivative of \( y = \sin^{-1} \left( \frac{1}{\sqrt{x + 1}} \right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x} (x + 1)} \]

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