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: Use the derivative of the inverse sine function
The derivative of \( \sin^{-1}(u) \) with respect to \( x \) 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 compute \( u^2 \):
\[ u^2 = \left(\frac{1}{\sqrt{x+1}}\right)^2 = \frac{1}{x+1} \]
Now, 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 \( u \):
\[ u = (x+1)^{-\frac{1}{2}} \]
Using the power rule:
\[ \frac{du}{dx} = -\frac{1}{2}(x+1)^{-\frac{3}{2}} \cdot (1) = -\frac{1}{2(x+1)^{\frac{3}{2}}} \]

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

### Step 5: Simplify the expression
Now simplify \( \frac{1}{\sqrt{\frac{x}{x+1}}} \):
\[ \frac{1}{\sqrt{\frac{x}{x+1}}} = \frac{\sqrt{x+1}}{\sqrt{x}} \]
Thus,
\[ \frac{dy}{dx} = \frac{\sqrt{x+1}}{\sqrt{x}} \cdot \left(-\frac{1}{2(x+1)^{\frac{3}{2}}}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x}\cdot(x+1)} \]

### Final Answer
Therefore, the derivative is:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x}\cdot(x+1)} \]

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