`sec^-1((1+tan ^2x)/(1-tan ^2 x))`

Answer

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

### Step 1: Simplify the Inner Function
We start with the expression inside the secant inverse function:
\[ \frac{1 + \tan^2 x}{1 - \tan^2 x} \]
Using the identity \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \), we can rewrite this as:
\[ \frac{1 + \frac{\sin^2 x}{\cos^2 x}}{1 - \frac{\sin^2 x}{\cos^2 x}} = \frac{\frac{\cos^2 x + \sin^2 x}{\cos^2 x}}{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x - \sin^2 x} \]
Since \( \cos^2 x + \sin^2 x = 1 \), we have:
\[ \frac{1}{\cos^2 x - \sin^2 x} \]

### Step 2: Further Simplify the Denominator
Now, we can express \( \cos^2 x - \sin^2 x \) using the double angle identity:
\[ \cos^2 x - \sin^2 x = \cos 2x \]
Thus, we can rewrite our expression as:
\[ \frac{1}{\cos 2x} \]
This gives us:
\[ \sec 2x \]

### Step 3: Rewrite the Original Function
Now substituting back into our original function:
\[ y = \sec^{-1}(\sec 2x) \]
Since the secant inverse and secant functions are inverses of each other, we have:
\[ y = 2x \]

### Step 4: Differentiate with Respect to \( x \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x) = 2 \]

### Final Answer
Thus, the derivative of the given expression is:
\[ \frac{dy}{dx} = 2 \]

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