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

Options

A. `x^(2)e^(x^(2))tane^(x^(2))`
B. `e^(x^(2))tane^(x^(2))`
C. `2xe^(x^(2))tane^(x^(2))`
D. none of these

Answer

To find the derivative of the function \( y = \log(\sec(e^{x^2})) \), we will apply the chain rule and the properties of logarithmic differentiation. Let's go through the steps:

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

### Step 2: Differentiate \( u = \sec(e^{x^2}) \)
Next, we need to find \( \frac{du}{dx} \). The derivative of \( \sec(v) \) is \( \sec(v) \tan(v) \cdot \frac{dv}{dx} \). Here, \( v = e^{x^2} \).

Thus, we have:
\[ \frac{du}{dx} = \sec(e^{x^2}) \tan(e^{x^2}) \cdot \frac{d}{dx}(e^{x^2}) \]

### Step 3: Differentiate \( e^{x^2} \)
Now we differentiate \( e^{x^2} \):
\[ \frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x \]

### Step 4: Substitute back into \( \frac{du}{dx} \)
Now we can substitute this back into our expression for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \sec(e^{x^2}) \tan(e^{x^2}) \cdot (2x e^{x^2}) \]

### Step 5: Combine everything to find \( \frac{dy}{dx} \)
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the 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 Result
Thus, the derivative is:
\[ \frac{dy}{dx} = 2x \tan(e^{x^2}) e^{x^2} \]