Differentiate the following w.r.t. x :
`e^(sin^(-1)(x+1))`

Answer

To differentiate the function \( y = e^{\sin^{-1}(x+1)} \) with respect to \( x \), we will use the chain rule and the derivative of the inverse sine function.

### Step-by-step solution:

1. **Identify the outer and inner functions**:
- The outer function is \( e^u \) where \( u = \sin^{-1}(x+1) \).
- The inner function is \( u = \sin^{-1}(x+1) \).

2. **Differentiate the outer function**:
- The derivative of \( e^u \) with respect to \( u \) is \( e^u \).
- Therefore, \( \frac{dy}{du} = e^{\sin^{-1}(x+1)} \).

3. **Differentiate the inner function**:
- The derivative of \( \sin^{-1}(v) \) with respect to \( v \) is \( \frac{1}{\sqrt{1 - v^2}} \).
- Here, \( v = x + 1 \), so we need to differentiate \( v \) with respect to \( x \):
\[ \frac{dv}{dx} = 1 \]
- Now, applying the chain rule:
\[ \frac{du}{dx} = \frac{d}{dx}(\sin^{-1}(x+1)) = \frac{1}{\sqrt{1 - (x+1)^2}} \cdot \frac{dv}{dx} = \frac{1}{\sqrt{1 - (x+1)^2}} \cdot 1 = \frac{1}{\sqrt{1 - (x+1)^2}} \]

4. **Combine the derivatives using the chain rule**:
- Now, we can find \( \frac{dy}{dx} \) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^{\sin^{-1}(x+1)} \cdot \frac{1}{\sqrt{1 - (x+1)^2}} \]

5. **Final expression**:
- Thus, the derivative of \( y = e^{\sin^{-1}(x+1)} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{e^{\sin^{-1}(x+1)}}{\sqrt{1 - (x+1)^2}} \]