Find f'(x) when
`f(x)=sin^(-1)(1/sqrt(x+1))`.

Answer

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

### Step 1: Rewrite the Function
We start with the function:
\[ f(x) = \sin^{-1}\left(\frac{1}{\sqrt{x+1}}\right) \]

### Step 2: Use the Chain Rule
To differentiate \( f(x) \), we will use the chain rule. The derivative of \( \sin^{-1}(u) \) is given by:
\[ \frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \]
where \( u = \frac{1}{\sqrt{x+1}} \).

### Step 3: Differentiate the Inner Function
Now we need to differentiate \( u = \frac{1}{\sqrt{x+1}} \):
\[ u = (x+1)^{-1/2} \]
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: Apply the Chain Rule
Now we can apply the chain rule:
\[ f'(x) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
Substituting \( u \) and \( \frac{du}{dx} \):
\[ f'(x) = \frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{x+1}}\right)^2}} \cdot \left(-\frac{1}{2\sqrt{(x+1)^3}}\right) \]

### Step 5: Simplify the Expression
Now simplify \( 1 - u^2 \):
\[ 1 - u^2 = 1 - \frac{1}{x+1} = \frac{(x+1) - 1}{x+1} = \frac{x}{x+1} \]
Thus,
\[ \sqrt{1 - u^2} = \sqrt{\frac{x}{x+1}} = \frac{\sqrt{x}}{\sqrt{x+1}} \]

### Step 6: Substitute Back into the Derivative
Now substituting back into the derivative:
\[ f'(x) = \frac{1}{\frac{\sqrt{x}}{\sqrt{x+1}}} \cdot \left(-\frac{1}{2\sqrt{(x+1)^3}}\right) \]
This simplifies to:
\[ f'(x) = -\frac{\sqrt{x+1}}{2\sqrt{x} \cdot (x+1)^{3/2}} = -\frac{1}{2\sqrt{x} (x+1)} \]

### Final Result
Thus, the derivative is:
\[ f'(x) = -\frac{1}{2\sqrt{x}(x+1)} \]

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