`cos^(-1)((1-x^(2n))/(1+x^(2n)))`

Answer

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

### Step 1: Define the Function
Let
\[ y = \cos^{-1}\left(\frac{1 - x^{2n}}{1 + x^{2n}}\right) \]

### Step 2: Use a Trigonometric Substitution
To simplify the differentiation, we can set:
\[ x^n = \tan(\theta) \]
This implies:
\[ \theta = \tan^{-1}(x^n) \]

### Step 3: Rewrite the Function in Terms of \(\theta\)
Substituting \( \tan(\theta) \) into the function gives:
\[ y = \cos^{-1}\left(\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}\right) \]
Using the identity \( \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta) \), we can rewrite \( y \) as:
\[ y = \cos^{-1}(\cos(2\theta)) \]

### Step 4: Simplify the Expression
Since \( \cos^{-1}(\cos(2\theta)) = 2\theta \) (for \( \theta \) in the appropriate range), we have:
\[ y = 2\theta \]
Substituting back for \(\theta\):
\[ y = 2\tan^{-1}(x^n) \]

### Step 5: Differentiate with Respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x^n)) \]
Using the chain rule, we find:
\[ \frac{d}{dx}(\tan^{-1}(x^n)) = \frac{1}{1 + (x^n)^2} \cdot \frac{d}{dx}(x^n) \]
The derivative of \( x^n \) is:
\[ \frac{d}{dx}(x^n) = n x^{n-1} \]
Thus, we have:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^{2n}} \cdot n x^{n-1} \]

### Step 6: Final Expression
Combining everything, we get:
\[ \frac{dy}{dx} = \frac{2n x^{n-1}}{1 + x^{2n}} \]

### Summary
The derivative of \( y = \cos^{-1}\left(\frac{1 - x^{2n}}{1 + x^{2n}}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{2n x^{n-1}}{1 + x^{2n}} \]