`sin^(-1)"(x)/(1+x)`

Answer

To differentiate the function \( y = \frac{\sin^{-1}(x)}{1+x} \) with respect to \( x \), we will use the quotient rule of differentiation. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = \sin^{-1}(x) \) and \( v = 1+x \).

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = \sin^{-1}(x) \)
- \( v = 1 + x \)

### Step 2: Differentiate \( u \) and \( v \)

Now we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

1. The derivative of \( u = \sin^{-1}(x) \) is:
\[ \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

2. The derivative of \( v = 1 + x \) is:
\[ \frac{dv}{dx} = 1 \]

### Step 3: Apply the Quotient Rule

Now, we can apply the quotient rule:

\[ \frac{dy}{dx} = \frac{(1+x) \cdot \frac{1}{\sqrt{1 - x^2}} - \sin^{-1}(x) \cdot 1}{(1+x)^2} \]

### Step 4: Simplify the Expression

Substituting the derivatives into the quotient rule formula gives us:

\[ \frac{dy}{dx} = \frac{(1+x) \cdot \frac{1}{\sqrt{1 - x^2}} - \sin^{-1}(x)}{(1+x)^2} \]

### Step 5: Final Expression

Thus, the derivative of \( y = \frac{\sin^{-1}(x)}{1+x} \) is:

\[ \frac{dy}{dx} = \frac{(1+x)}{\sqrt{1 - x^2}(1+x)^2} - \frac{\sin^{-1}(x)}{(1+x)^2} \]

This can be further simplified, but this is the derivative in its current form.