If ` y= log sece^(x^(2)) ,then *(dy)/(dx) `

Options

A. ` xe^(x^(2)) tan e^(x^(2))`
B. ` 2 xe^(x^(2)) tan e^(x^(2))`
C. ` - xe^(x^(2)) tan e^(x^(2))`
D. ` - 2xe^(x^(2)) tan e^(x^(2))`

Answer

To find the derivative of the function \( y = \log(\sec(e^{x^2})) \), we will use the chain rule and the properties of logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate the logarithmic function
Using the chain rule, the derivative of \( y = \log(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sec(e^{x^2}) \).

### Step 2: Find \( \frac{du}{dx} \)
Now we need to differentiate \( u = \sec(e^{x^2}) \). The derivative of \( \sec(v) \) is \( \sec(v) \tan(v) \cdot \frac{dv}{dx} \), where \( v = e^{x^2} \).

### Step 3: Differentiate \( v = e^{x^2} \)
Next, we find \( \frac{dv}{dx} \) where \( v = e^{x^2} \):
\[ \frac{dv}{dx} = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot (2x) \]

### Step 4: Combine the derivatives
Now substituting back into the derivative of \( u \):
\[ \frac{du}{dx} = \sec(e^{x^2}) \tan(e^{x^2}) \cdot \frac{dv}{dx} = \sec(e^{x^2}) \tan(e^{x^2}) \cdot (2x e^{x^2}) \]

### Step 5: Substitute \( \frac{du}{dx} \) back into \( \frac{dy}{dx} \)
Now we can substitute \( \frac{du}{dx} \) back into our expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{\sec(e^{x^2})} \cdot \left( \sec(e^{x^2}) \tan(e^{x^2}) \cdot (2x e^{x^2}) \right) \]

### Step 6: Simplify the expression
The \( \sec(e^{x^2}) \) terms cancel out:
\[ \frac{dy}{dx} = \tan(e^{x^2}) \cdot (2x e^{x^2}) \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 2x e^{x^2} \tan(e^{x^2}) \]

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