The function `y=tan^(-1) x-x`

Options

A. is always decreasing
B. is always increasing
C. first increases and then decreases
D. first decreases and then increases

Explanation

To determine the nature of the function \( y = \tan^{-1} x - x \), we need to analyze its derivative. Here are the steps to solve the problem:

### Step 1: Differentiate the function
We start with the function:
\[ y = \tan^{-1} x - x \]
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1} x) - \frac{d}{dx}(x) \]
Using the derivative of \( \tan^{-1} x \), which is \( \frac{1}{1+x^2} \), we have:
\[ \frac{dy}{dx} = \frac{1}{1+x^2} - 1 \]

### Step 2: Simplify the derivative
Now, we simplify the expression for the derivative:
\[ \frac{dy}{dx} = \frac{1}{1+x^2} - \frac{1+x^2}{1+x^2} = \frac{1 - (1 + x^2)}{1 + x^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1 - 1 - x^2}{1 + x^2} = \frac{-x^2}{1 + x^2} \]

### Step 3: Analyze the sign of the derivative
The expression for the derivative is:
\[ \frac{dy}{dx} = \frac{-x^2}{1 + x^2} \]
Since \( x^2 \) is always non-negative for all real \( x \) (i.e., \( x^2 \geq 0 \)), it follows that \( -x^2 \) is always non-positive (i.e., \( -x^2 \leq 0 \)).

Furthermore, the denominator \( 1 + x^2 \) is always positive for all real \( x \) (i.e., \( 1 + x^2 > 0 \)).

Thus, the derivative \( \frac{dy}{dx} \) is always non-positive:
\[ \frac{dy}{dx} \leq 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 4: Conclusion about the function
Since the derivative is always non-positive, this indicates that the function \( y = \tan^{-1} x - x \) is always decreasing for all values of \( x \) in the real numbers.

### Final Answer
The function \( y = \tan^{-1} x - x \) is always decreasing.

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