`cot^(-1)(sqrt(1+x^2)-x)`

Answer

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

### Step 1: Identify the function to differentiate
We have:
\[ y = \cot^{-1}(u) \quad \text{where } u = \sqrt{1+x^2} - x \]

### Step 2: Differentiate the outer function
The derivative of \( \cot^{-1}(u) \) is given by:
\[ \frac{dy}{du} = -\frac{1}{1 + u^2} \]

### Step 3: Differentiate the inner function \( u \)
Now, we need to differentiate \( u \):
\[ u = \sqrt{1+x^2} - x \]
To differentiate \( u \), we will use the chain rule. The derivative of \( \sqrt{1+x^2} \) is:
\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{1}{2\sqrt{1+x^2}} \cdot (2x) = \frac{x}{\sqrt{1+x^2}} \]
The derivative of \( -x \) is simply \( -1 \). Therefore, we have:
\[ \frac{du}{dx} = \frac{x}{\sqrt{1+x^2}} - 1 \]

### Step 4: Combine the derivatives using the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{1 + u^2} \cdot \left(\frac{x}{\sqrt{1+x^2}} - 1\right) \]

### Step 5: Substitute back for \( u \)
Next, we need to substitute back \( u = \sqrt{1+x^2} - x \) into the expression:
\[ 1 + u^2 = 1 + (\sqrt{1+x^2} - x)^2 \]
Calculating \( u^2 \):
\[ u^2 = (\sqrt{1+x^2} - x)^2 = (1+x^2) - 2x\sqrt{1+x^2} + x^2 = 1 + 2x^2 - 2x\sqrt{1+x^2} \]
Thus,
\[ 1 + u^2 = 2 + 2x^2 - 2x\sqrt{1+x^2} \]

### Step 6: Final expression for the derivative
Putting it all together, we have:
\[ \frac{dy}{dx} = -\frac{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot \left(\frac{x}{\sqrt{1+x^2}} - 1\right) \]

### Final Answer
Thus, the derivative of \( y = \cot^{-1}(\sqrt{1+x^2} - x) \) is:
\[ \frac{dy}{dx} = -\frac{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot \left(\frac{x}{\sqrt{1+x^2}} - 1\right) \]