Differentiate the function `y =sec (x^2+2)` w.r.t. x is :

Options

A. `2x sec(x^2 +2 ) tan (x^2+2)`
B. `sec(x^2+2) tan (x^2+2)`
C. `2x sec^2 (x^2+2)`
D. `2x tan^2(x^2+2)`

Answer

To differentiate the function \( y = \sec(x^2 + 2) \) with respect to \( x \), we will use the chain rule. Here are the steps to find the derivative:

### Step 1: Identify the outer and inner functions
The function can be expressed as:
- Outer function: \( u = \sec(v) \)
- Inner function: \( v = x^2 + 2 \)

### Step 2: Differentiate the outer function
The derivative of \( \sec(v) \) with respect to \( v \) is:
\[ \frac{du}{dv} = \sec(v) \tan(v) \]

### Step 3: Differentiate the inner function
The derivative of \( v = x^2 + 2 \) with respect to \( x \) is:
\[ \frac{dv}{dx} = 2x \]

### Step 4: Apply the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \sec(v) \tan(v) \cdot 2x \]

### Step 5: Substitute back the inner function
Now substitute \( v = x^2 + 2 \) back into the equation:
\[ \frac{dy}{dx} = \sec(x^2 + 2) \tan(x^2 + 2) \cdot 2x \]

### Final Answer
Thus, the derivative of \( y = \sec(x^2 + 2) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 2x \sec(x^2 + 2) \tan(x^2 + 2) \]

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