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

Answer

To differentiate the function \( y = e^{\cos^{-1}(x+1)} \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = e^{\cos^{-1}(x+1)} \]

### Step 2: Apply the chain rule
To differentiate \( y \) with respect to \( x \), we can use the chain rule. The derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \), where \( u = \cos^{-1}(x+1) \).

Thus, we have:
\[ \frac{dy}{dx} = e^{\cos^{-1}(x+1)} \cdot \frac{d}{dx}(\cos^{-1}(x+1)) \]

### Step 3: Differentiate \( \cos^{-1}(x+1) \)
The derivative of \( \cos^{-1}(u) \) is given by:
\[ \frac{d}{du}(\cos^{-1}(u)) = -\frac{1}{\sqrt{1-u^2}} \]
where \( u = x + 1 \). Therefore, we need to apply the chain rule again for \( u = x + 1 \):
\[ \frac{d}{dx}(\cos^{-1}(x+1)) = -\frac{1}{\sqrt{1-(x+1)^2}} \cdot \frac{d}{dx}(x+1) \]
Since \( \frac{d}{dx}(x+1) = 1 \), we have:
\[ \frac{d}{dx}(\cos^{-1}(x+1)) = -\frac{1}{\sqrt{1-(x+1)^2}} \]

### Step 4: Substitute back into the derivative
Now substituting this back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = e^{\cos^{-1}(x+1)} \cdot \left(-\frac{1}{\sqrt{1-(x+1)^2}}\right) \]

### Step 5: Simplify the expression
Thus, we can write:
\[ \frac{dy}{dx} = -\frac{e^{\cos^{-1}(x+1)}}{\sqrt{1-(x+1)^2}} \]

### Final Result
The derivative of \( y = e^{\cos^{-1}(x+1)} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{e^{\cos^{-1}(x+1)}}{\sqrt{1-(x+1)^2}} \]