If ` Y= log [Sec (e^(x^2))] ` then ` (dy)/(dx)` is equal to

Options

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

Answer

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

### Step 1: Apply the Logarithmic Identity
We can use the property of logarithms that states \( \log(a^b) = b \log(a) \). In this case, we can express \( Y \) as:
\[ Y = \log(\sec(e^{x^2})) = \log(\sec(e^{x^2})) = \log(\sec(u)) \text{ where } u = e^{x^2} \]

### Step 2: Differentiate Using the Chain Rule
To differentiate \( Y \) with respect to \( x \), we will apply the chain rule:
\[ \frac{dY}{dx} = \frac{dY}{du} \cdot \frac{du}{dx} \]

### Step 3: Differentiate \( Y \) with Respect to \( u \)
Now we need to differentiate \( Y \) with respect to \( u \):
\[ \frac{dY}{du} = \frac{d}{du}(\log(\sec(u))) = \frac{1}{\sec(u)} \cdot \frac{d}{du}(\sec(u)) \]
The derivative of \( \sec(u) \) is \( \sec(u) \tan(u) \), so:
\[ \frac{dY}{du} = \frac{1}{\sec(u)} \cdot \sec(u) \tan(u) = \tan(u) \]

### Step 4: Differentiate \( u \) with Respect to \( x \)
Next, we differentiate \( u = e^{x^2} \):
\[ \frac{du}{dx} = \frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x \]

### Step 5: Combine the Results
Now we can combine the derivatives:
\[ \frac{dY}{dx} = \tan(u) \cdot \frac{du}{dx} = \tan(e^{x^2}) \cdot (2x e^{x^2}) \]

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

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