Find the derivative of the following function with respect to 'x'
`cot^(-1)(sqrt(1+x^2)+x)`

Answer

To find the derivative of the function \( y = \cot^{-1}(\sqrt{1+x^2} + x) \), we will follow these steps:

### Step 1: Rewrite the Function
Let:
\[ y = \cot^{-1}(\sqrt{1+x^2} + x) \]

### Step 2: Differentiate Using the Chain Rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( \cot^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = -\frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \sqrt{1+x^2} + x \).

### Step 3: Find \( \frac{du}{dx} \)
Now, we need to find \( \frac{du}{dx} \):
\[ u = \sqrt{1+x^2} + x \]
Differentiating \( u \):
\[ \frac{du}{dx} = \frac{d}{dx}(\sqrt{1+x^2}) + \frac{d}{dx}(x) \]
Using the derivative of \( \sqrt{1+x^2} \):
\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{x}{\sqrt{1+x^2}} \]
Thus,
\[ \frac{du}{dx} = \frac{x}{\sqrt{1+x^2}} + 1 \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the Derivative Formula
Now we substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = -\frac{1}{1 + (\sqrt{1+x^2} + x)^2} \cdot \left( \frac{x}{\sqrt{1+x^2}} + 1 \right) \]

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

### Final Derivative
Putting it all together:
\[ \frac{dy}{dx} = -\frac{1}{2 + x^2 + 2x\sqrt{1+x^2}} \cdot \left( \frac{x}{\sqrt{1+x^2}} + 1 \right) \]