`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: Rewrite the function
We can express the function in a more manageable form. Recognizing that:
\[ y = \cos^{-1}\left(\frac{1-x}{1+x}\right) \]
can be rewritten using the identity for the inverse cosine function.

### Step 2: Use the derivative of the inverse cosine
The derivative of \( y = \cos^{-1}(u) \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{1-x}{1+x} \).

### Step 3: Differentiate \( u \)
Now, we need to find \( \frac{du}{dx} \):
\[ u = \frac{1-x}{1+x} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1 - x - 1 + x}{(1+x)^2} = \frac{-2}{(1+x)^2} \]

### Step 4: Calculate \( u^2 \)
Next, we need to calculate \( 1 - u^2 \):
\[ u^2 = \left(\frac{1-x}{1+x}\right)^2 = \frac{(1-x)^2}{(1+x)^2} \]
Thus,
\[ 1 - u^2 = 1 - \frac{(1-x)^2}{(1+x)^2} = \frac{(1+x)^2 - (1-x)^2}{(1+x)^2} \]
Calculating the numerator:
\[ (1+x)^2 - (1-x)^2 = (1 + 2x + x^2) - (1 - 2x + x^2) = 4x \]
So,
\[ 1 - u^2 = \frac{4x}{(1+x)^2} \]

### Step 5: Substitute into the derivative formula
Now we can substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{\frac{4x}{(1+x)^2}}} \cdot \left(-\frac{2}{(1+x)^2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{2}{(1+x)^2 \cdot \sqrt{\frac{4x}{(1+x)^2}}} = \frac{2}{(1+x)^2} \cdot \frac{(1+x)}{2\sqrt{x}} = \frac{1}{(1+x)\sqrt{x}} \]

### Final Answer
Thus, the derivative of the function \( y = \cos^{-1}\left(\frac{1-x}{1+x}\right) \) is:
\[ \frac{dy}{dx} = \frac{1}{(1+x)\sqrt{x}} \]