Differentiate `sin^(-1)\ (1)/(sqrt(x+1))`

Explanation

To differentiate the function \( y = \sin^{-1} \left( \frac{1}{\sqrt{x+1}} \right) \), we will follow these steps:

### Step 1: Identify the function
Let:
\[ y = \sin^{-1} \left( \frac{1}{\sqrt{x+1}} \right) \]

### Step 2: Differentiate using the chain rule
Using 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 \( u \) and its derivative
First, we need to find \( u^2 \):
\[ u^2 = \left( \frac{1}{\sqrt{x+1}} \right)^2 = \frac{1}{x+1} \]
Then, we can find \( 1 - u^2 \):
\[ 1 - u^2 = 1 - \frac{1}{x+1} = \frac{x+1 - 1}{x+1} = \frac{x}{x+1} \]

Next, we differentiate \( u \):
\[ u = (x+1)^{-1/2} \]
Using the power rule:
\[ \frac{du}{dx} = -\frac{1}{2} (x+1)^{-3/2} \cdot \frac{d}{dx}(x+1) = -\frac{1}{2} (x+1)^{-3/2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now substituting \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} = \frac{1}{\sqrt{\frac{x}{x+1}}} \cdot \left(-\frac{1}{2} (x+1)^{-3/2}\right) \]

### Step 5: Simplify the expression
We can simplify \( \sqrt{1 - u^2} \):
\[ \sqrt{1 - u^2} = \sqrt{\frac{x}{x+1}} = \frac{\sqrt{x}}{\sqrt{x+1}} \]
Thus, substituting this back:
\[ \frac{dy}{dx} = \frac{\sqrt{x+1}}{\sqrt{x}} \cdot \left(-\frac{1}{2} (x+1)^{-3/2}\right) \]

### Step 6: Final expression for the derivative
This simplifies to:
\[ \frac{dy}{dx} = -\frac{1}{2} \cdot \frac{\sqrt{x+1}}{\sqrt{x} \cdot (x+1)^{3/2}} = -\frac{1}{2} \cdot \frac{1}{\sqrt{x} (x+1)} \]

Thus, the final derivative is:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x}(x+1)} \]