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

Options

A. ` (1)/(1+x^(2))`
B. ` (-1)/(1+x^(2))`
C. ` (2)/(1+x^(2))`
D. ` (-2)/(1+x^(2))`

Answer

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

### Step 1: Recognize the relationship
We know from trigonometric identities that:
\[ \frac{1 - x^2}{1 + x^2} = \cos(2\theta) \quad \text{where} \quad x = \tan(\theta) \]
Thus, we can express \( y \) in terms of \( \theta \):
\[ y = \cos^{-1}(\cos(2\theta)) = 2\theta \]

### Step 2: Express \( \theta \) in terms of \( x \)
Since \( x = \tan(\theta) \), we have:
\[ \theta = \tan^{-1}(x) \]
Substituting this back into our expression for \( y \):
\[ y = 2\tan^{-1}(x) \]

### Step 3: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]
Thus, we have:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{2}{1 + x^2} \]

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