The function `y=cot^(-1) x- log (x+sqrt(x^2+1))` is decreasing in

Options

A. `(-oo , 0)`
B. `(-oo,0)`
C. `(0,oo)`
D. `(-oo,oo)`

Answer

To determine where the function \( y = \cot^{-1}(x) - \log(x + \sqrt{x^2 + 1}) \) is decreasing, we need to find its derivative and analyze its sign.

### Step-by-Step Solution:

1. **Differentiate the Function**:
We start by differentiating the function \( y \):
\[ y' = \frac{d}{dx} \left( \cot^{-1}(x) \right) - \frac{d}{dx} \left( \log(x + \sqrt{x^2 + 1}) \right) \]

2. **Differentiate \( \cot^{-1}(x) \)**:
The derivative of \( \cot^{-1}(x) \) is:
\[ \frac{d}{dx} \left( \cot^{-1}(x) \right) = -\frac{1}{1 + x^2} \]

3. **Differentiate \( \log(x + \sqrt{x^2 + 1}) \)**:
Using the chain rule, we differentiate \( \log(u) \) where \( u = x + \sqrt{x^2 + 1} \):
\[ \frac{d}{dx} \left( \log(u) \right) = \frac{1}{u} \cdot \frac{du}{dx} \]
First, we find \( \frac{du}{dx} \):
\[ u = x + \sqrt{x^2 + 1} \implies \frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2 + 1}} \]
Thus,
\[ \frac{d}{dx} \left( \log(x + \sqrt{x^2 + 1}) \right) = \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} \]

4. **Combine Derivatives**:
Now, substituting back into the derivative of \( y \):
\[ y' = -\frac{1}{1 + x^2} - \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} \]

5. **Simplify the Expression**:
To simplify, we find a common denominator:
\[ y' = -\frac{1}{1 + x^2} - \frac{(1 + \frac{x}{\sqrt{x^2 + 1}})}{x + \sqrt{x^2 + 1}} \]
Combine the terms:
\[ y' = -\left( \frac{(x + \sqrt{x^2 + 1}) + (1 + \frac{x}{\sqrt{x^2 + 1}})(1 + x^2)}{(1 + x^2)(x + \sqrt{x^2 + 1})} \right) \]

6. **Analyze the Sign of \( y' \)**:
The expression for \( y' \) is negative for all \( x \) because both components are negative. Hence, \( y' < 0 \) for all \( x \).

7. **Conclusion**:
Since \( y' < 0 \) for all \( x \), the function \( y \) is decreasing in the interval \( (-\infty, \infty) \).

### Final Answer:
The function \( y = \cot^{-1}(x) - \log(x + \sqrt{x^2 + 1}) \) is decreasing in the interval \( (-\infty, \infty) \).