Differentiate the following w.r.t. as indicated :
`tan^(2)x" w.r.t. "sec^(2)(x^(2))`

Answer

To differentiate \( \tan^2 x \) with respect to \( \sec^2(x^2) \), we will use the chain rule. Let's denote:

- \( u = \tan^2 x \)
- \( v = \sec^2(x^2) \)

### Step 1: Differentiate \( u \) with respect to \( x \)

Using the chain rule for \( u \):

\[ \frac{du}{dx} = 2\tan x \cdot \sec^2 x \]

### Step 2: Differentiate \( v \) with respect to \( x \)

Using the chain rule for \( v \):

\[ \frac{dv}{dx} = 2\sec^2(x^2) \cdot \tan(x^2) \cdot 2x = 4x \sec^2(x^2) \tan(x^2) \]

### Step 3: Find \( \frac{du}{dv} \)

Using the formula \( \frac{du}{dv} = \frac{du/dx}{dv/dx} \):

\[ \frac{du}{dv} = \frac{2\tan x \cdot \sec^2 x}{4x \sec^2(x^2) \tan(x^2)} \]

### Step 4: Simplify the expression

We can simplify the expression:

\[ \frac{du}{dv} = \frac{2\tan x \cdot \sec^2 x}{4x \sec^2(x^2) \tan(x^2)} = \frac{\tan x \cdot \sec^2 x}{2x \sec^2(x^2) \tan(x^2)} \]

### Final Result

Thus, the derivative of \( \tan^2 x \) with respect to \( \sec^2(x^2) \) is:

\[ \frac{du}{dv} = \frac{\tan x \cdot \sec^2 x}{2x \sec^2(x^2) \tan(x^2)} \]