If `y=cos^(-1)((2x)/(1+x^(2)))`, then `(dy)/(dx)` is equal to

Options

A. `-(2)/(1+x^(2))" for all " |x|lt1`
B. `-(2)/(1+x^(2))" for all "|x|gt1`
C. `(2)/(1+x^(2))" for all " |x|lt1`
D. None of these

Answer

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

### Step 1: Substitute \( x = \tan(\theta) \)
Let \( x = \tan(\theta) \). Then, we have:
\[ \frac{2x}{1+x^2} = \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]
This is based on the double angle identity for sine.

### Step 2: Rewrite \( y \)
Now we can rewrite \( y \) as:
\[ y = \cos^{-1}(\sin(2\theta)) \]

### Step 3: Use the identity for cosine
Using the identity \( \cos^{-1}(\sin(2\theta)) = \frac{\pi}{2} - 2\theta \):
\[ y = \frac{\pi}{2} - 2\theta \]

### Step 4: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 2\theta\right) \]
Since \( \frac{\pi}{2} \) is a constant, its derivative is 0. Thus:
\[ \frac{dy}{dx} = -2 \frac{d\theta}{dx} \]

### Step 5: Find \( \frac{d\theta}{dx} \)
Recall that \( \theta = \tan^{-1}(x) \). Therefore, the derivative is:
\[ \frac{d\theta}{dx} = \frac{1}{1+x^2} \]

### Step 6: Substitute back into the derivative
Substituting \( \frac{d\theta}{dx} \) back into the equation for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1+x^2} = -\frac{2}{1+x^2} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{2}{1+x^2} \]