Differentiate the following w.r.t. x :
`(sec^(2)x)^(1//x)`

Answer

To differentiate the function \( y = (\sec^2 x)^{\frac{1}{x}} \) with respect to \( x \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process.

\[ \ln y = \ln\left((\sec^2 x)^{\frac{1}{x}}\right) \]

Using the property of logarithms, we can bring down the exponent:

\[ \ln y = \frac{1}{x} \ln(\sec^2 x) \]

### Step 2: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. We will use implicit differentiation on the left side and the product rule on the right side.

The left side becomes:

\[ \frac{1}{y} \frac{dy}{dx} \]

For the right side, we apply the product rule:

\[ \frac{d}{dx}\left(\frac{1}{x} \ln(\sec^2 x)\right) = \frac{d}{dx}\left(\frac{1}{x}\right) \ln(\sec^2 x) + \frac{1}{x} \frac{d}{dx}(\ln(\sec^2 x)) \]

The derivative of \( \frac{1}{x} \) is \( -\frac{1}{x^2} \).

### Step 3: Differentiate \( \ln(\sec^2 x) \)
Next, we differentiate \( \ln(\sec^2 x) \):

\[ \frac{d}{dx}(\ln(\sec^2 x)) = \frac{1}{\sec^2 x} \cdot \frac{d}{dx}(\sec^2 x) \]

Using the chain rule, we find \( \frac{d}{dx}(\sec^2 x) = 2 \sec^2 x \tan x \).

Thus,

\[ \frac{d}{dx}(\ln(\sec^2 x)) = \frac{2 \sec^2 x \tan x}{\sec^2 x} = 2 \tan x \]

### Step 4: Substitute back into the differentiation equation
Now we substitute back into our differentiation equation:

\[ \frac{1}{y} \frac{dy}{dx} = -\frac{1}{x^2} \ln(\sec^2 x) + \frac{1}{x} \cdot 2 \tan x \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \) gives us:

\[ \frac{dy}{dx} = y \left(-\frac{1}{x^2} \ln(\sec^2 x) + \frac{2 \tan x}{x}\right) \]

### Step 6: Substitute \( y \) back into the equation
Recalling that \( y = (\sec^2 x)^{\frac{1}{x}} \):

\[ \frac{dy}{dx} = (\sec^2 x)^{\frac{1}{x}} \left(-\frac{1}{x^2} \ln(\sec^2 x) + \frac{2 \tan x}{x}\right) \]

### Final Answer
Thus, the derivative of \( y = (\sec^2 x)^{\frac{1}{x}} \) with respect to \( x \) is:

\[ \frac{dy}{dx} = (\sec^2 x)^{\frac{1}{x}} \left(-\frac{1}{x^2} \ln(\sec^2 x) + \frac{2 \tan x}{x}\right) \]