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 \( \frac{dy}{dx} \) for the function \( y = \cos^{-1}\left(\frac{2x}{1+x^2}\right) \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = \cos^{-1}\left(\frac{2x}{1+x^2}\right) \]

### Step 2: Use the trigonometric identity
Recognize that \( \frac{2x}{1+x^2} \) can be expressed in terms of the sine of an angle. Specifically, we can set \( x = \tan(\theta) \). Then:
\[ \frac{2x}{1+x^2} = \sin(2\theta) \]
Thus, we can rewrite \( y \):
\[ y = \cos^{-1}(\sin(2\theta)) \]

### Step 3: Simplify using the complementary angle
Using the identity \( \cos^{-1}(\sin(2\theta)) = \frac{\pi}{2} - 2\theta \):
\[ y = \frac{\pi}{2} - 2\theta \]

### Step 4: Substitute back for \( \theta \)
Since \( \theta = \tan^{-1}(x) \), we have:
\[ y = \frac{\pi}{2} - 2\tan^{-1}(x) \]

### Step 5: Differentiate with respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0 - 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \]
Thus,
\[ \frac{dy}{dx} = -2 \cdot \frac{1}{1+x^2} = -\frac{2}{1+x^2} \]

### Step 6: Conclusion
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = -\frac{2}{1+x^2} \]