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: Differentiate the Inverse Sine Function
First, we need to differentiate \( y \) with respect to \( x \). The derivative of \( \sin^{-1}(u) \) is given by the formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{1}{\sqrt{x+1}} \).

### Step 2: Find \( u \) and its Derivative
Now, we need to find \( \frac{du}{dx} \):
\[ u = \frac{1}{\sqrt{x+1}} \implies u = (x+1)^{-1/2} \]
Using the power rule for differentiation:
\[ \frac{du}{dx} = -\frac{1}{2}(x+1)^{-3/2} \cdot \frac{d}{dx}(x+1) = -\frac{1}{2}(x+1)^{-3/2} \]

### Step 3: Substitute \( u \) into the Derivative Formula
Next, we substitute \( u \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{x+1}}\right)^2}} \cdot \left(-\frac{1}{2}(x+1)^{-3/2}\right) \]

### Step 4: Simplify the Expression
Now, simplify \( \sqrt{1 - \left(\frac{1}{\sqrt{x+1}}\right)^2} \):
\[ 1 - \left(\frac{1}{\sqrt{x+1}}\right)^2 = 1 - \frac{1}{x+1} = \frac{(x+1) - 1}{x+1} = \frac{x}{x+1} \]
Thus,
\[ \sqrt{1 - \left(\frac{1}{\sqrt{x+1}}\right)^2} = \sqrt{\frac{x}{x+1}} = \frac{\sqrt{x}}{\sqrt{x+1}} \]

### Step 5: Substitute Back into the Derivative
Now substitute this back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\frac{\sqrt{x}}{\sqrt{x+1}}} \cdot \left(-\frac{1}{2}(x+1)^{-3/2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sqrt{x+1}}{2\sqrt{x}} \cdot (x+1)^{-3/2} \]

### Step 6: Final Simplification
Finally, we simplify:
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x}(x+1)} \]

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