If `y=cos^(-1)((1-x^(2))/(1+x^(2))),` then
Statement I `(dy)/(dx)=(2)/(1+x^(2))` for `x"inR`
Statement II `cos^(-1)((1-x^(2))/(1+x^(2)))={(2tan^(-1)x,,xge0,),(-2tan^(-1)x,,xlt0,):}`

Options

A. Both statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
C. Statement I is correct but Statement II is incorrect
D. Statement II is correct but Statement I is incorrect.

Answer

To solve the problem, we start with the given function:

\[ y = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]

### Step 1: Differentiate the function
To find \(\frac{dy}{dx}\), we will use the chain rule. The derivative of \(\cos^{-1}(u)\) is given by:

\[ \frac{d}{du}(\cos^{-1}(u)) = -\frac{1}{\sqrt{1 - u^2}} \]

Let \( u = \frac{1 - x^2}{1 + x^2} \). We need to find \(\frac{du}{dx}\).

### Step 2: Differentiate \(u\)
Using the quotient rule, we have:

\[ u = \frac{1 - x^2}{1 + x^2} \]

Let \( f(x) = 1 - x^2 \) and \( g(x) = 1 + x^2 \). Then, by the quotient rule:

\[ \frac{du}{dx} = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \]

Calculating \(f'(x)\) and \(g'(x)\):

- \(f'(x) = -2x\)
- \(g'(x) = 2x\)

Now substituting back:

\[ \frac{du}{dx} = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]

Simplifying the numerator:

\[ = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 + x^2)^2} = \frac{-4x}{(1 + x^2)^2} \]

### Step 3: Substitute back into the derivative of \(y\)
Now substituting \(u\) and \(\frac{du}{dx}\) into the derivative of \(y\):

\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]

### Step 4: Calculate \(1 - u^2\)
We need to calculate \(1 - u^2\):

\[ u^2 = \left(\frac{1 - x^2}{1 + x^2}\right)^2 = \frac{(1 - x^2)^2}{(1 + x^2)^2} \]

Thus,

\[ 1 - u^2 = 1 - \frac{(1 - x^2)^2}{(1 + x^2)^2} = \frac{(1 + x^2)^2 - (1 - x^2)^2}{(1 + x^2)^2} \]

Calculating the numerator:

\[ (1 + x^2)^2 - (1 - x^2)^2 = (1 + 2x^2 + x^4) - (1 - 2x^2 + x^4) = 4x^2 \]

So,

\[ 1 - u^2 = \frac{4x^2}{(1 + x^2)^2} \]

### Step 5: Substitute back into \(\frac{dy}{dx}\)
Now substituting \(1 - u^2\):

\[ \frac{dy}{dx} = -\frac{1}{\sqrt{\frac{4x^2}{(1 + x^2)^2}}} \cdot \frac{-4x}{(1 + x^2)^2} \]

Calculating \(\sqrt{1 - u^2}\):

\[ \sqrt{1 - u^2} = \frac{2|x|}{1 + x^2} \]

Thus,

\[ \frac{dy}{dx} = \frac{4x}{(1 + x^2)^2} \cdot \frac{1 + x^2}{2|x|} = \frac{2}{1 + x^2} \]

### Conclusion
Thus, we have:

\[ \frac{dy}{dx} = \frac{2}{1 + x^2} \quad \text{for } x \geq 0 \]
\[ \frac{dy}{dx} = -\frac{2}{1 + x^2} \quad \text{for } x < 0 \]