Differentiate the following w.r.t. x :
`cos^(-1)((1-x^(2))/(1+x^(2))),0ltxlt1`

Answer

To differentiate the function \( y = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \) with respect to \( x \), we will follow these steps:

### Step 1: Substitute \( x \) with \( \tan(\theta) \)
Let \( x = \tan(\theta) \). Then, we can express the function in terms of \( \theta \):

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

### Step 2: Simplify the expression
Using the identity \( \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta) \), we can rewrite \( y \):

\[ y = \cos^{-1}(\cos(2\theta)) \]

### Step 3: Determine the value of \( y \)
Since \( \cos^{-1}(\cos(2\theta)) = 2\theta \) (for \( 0 < \theta < \frac{\pi}{4} \)), we have:

\[ y = 2\theta \]

### Step 4: Differentiate \( y \) with respect to \( \theta \)
Now, we differentiate \( y \) with respect to \( \theta \):

\[ \frac{dy}{d\theta} = 2 \]

### Step 5: Differentiate \( \theta \) with respect to \( x \)
Next, we need to find \( \frac{d\theta}{dx} \). Since \( \theta = \tan^{-1}(x) \), we have:

\[ \frac{d\theta}{dx} = \frac{1}{1 + x^2} \]

### Step 6: Apply the chain rule
Now, we apply the chain rule to find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} = 2 \cdot \frac{1}{1 + x^2} \]

### Final Result
Thus, the derivative of the given function with respect to \( x \) is:

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

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