`cos^(-1)((1-x)/(1+x))`

Answer

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

### Step 1: Set the function
Let
\[ y = \cos^{-1}\left(\frac{1-x}{1+x}\right) \]

### Step 2: Use a trigonometric substitution
We can use the substitution \( x = \tan^2(\theta) \). Then we have:
\[ 1 - x = 1 - \tan^2(\theta) = \frac{\cos(2\theta)}{\cos^2(\theta)} \]
\[ 1 + x = 1 + \tan^2(\theta) = \frac{1}{\cos^2(\theta)} \]
Thus,
\[ \frac{1-x}{1+x} = \frac{\cos(2\theta)}{1} = \cos(2\theta) \]

### Step 3: Rewrite the function
Now we can rewrite \( y \):
\[ y = \cos^{-1}(\cos(2\theta)) = 2\theta \]

### Step 4: Relate \( \theta \) back to \( x \)
Since \( x = \tan^2(\theta) \), we have:
\[ \tan(\theta) = \sqrt{x} \implies \theta = \tan^{-1}(\sqrt{x}) \]
Thus,
\[ y = 2\tan^{-1}(\sqrt{x}) \]

### Step 5: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(\sqrt{x})) \]
Using the chain rule:
\[ \frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1+u^2} \cdot \frac{du}{dx} \]
where \( u = \sqrt{x} \) and \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \).

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

### Final Answer
The derivative of \( y = \cos^{-1}\left(\frac{1-x}{1+x}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x}(1+x)} \]

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