If `y(x) = x^x , x gt 0` then `y"(2) – 2y'(2)` is equal to

Options

A. `4 (log_(e)2)^2 – 2`
B. `8 log_(e)2 – 2`
C. `4 (log_(e)2)^2+ 2`
D. `4 log_(e)2 + 2`

Answer

To solve the problem, we need to find \( y''(2) - 2y'(2) \) for the function \( y(x) = x^x \).

### Step 1: Find \( y'(x) \)

We start by differentiating \( y(x) = x^x \). To do this, we can use logarithmic differentiation.

1. Take the natural logarithm of both sides:
\[ \ln y = \ln(x^x) = x \ln x \]

2. Differentiate both sides with respect to \( x \):
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

3. Multiply through by \( y \):
\[ y' = y(\ln x + 1) = x^x(\ln x + 1) \]

### Step 2: Find \( y'(2) \)

Now we substitute \( x = 2 \):
\[ y'(2) = 2^2(\ln 2 + 1) = 4(\ln 2 + 1) \]

### Step 3: Find \( y''(x) \)

Next, we differentiate \( y'(x) \) to find \( y''(x) \). We will use the product rule:
\[ y' = x^x(\ln x + 1) \]

Using the product rule:
\[ y'' = \frac{d}{dx}(x^x) \cdot (\ln x + 1) + x^x \cdot \frac{d}{dx}(\ln x + 1) \]

We already found \( \frac{d}{dx}(x^x) = x^x(\ln x + 1) \), and:
\[ \frac{d}{dx}(\ln x + 1) = \frac{1}{x} \]

Thus:
\[ y'' = x^x(\ln x + 1)(\ln x + 1) + x^x \cdot \frac{1}{x} \]
\[ = x^x(\ln x + 1)^2 + x^{x-1} \]

### Step 4: Find \( y''(2) \)

Now we substitute \( x = 2 \):
\[ y''(2) = 2^2(\ln 2 + 1)^2 + 2^{2-1} = 4(\ln 2 + 1)^2 + 2 \]

### Step 5: Calculate \( y''(2) - 2y'(2) \)

Now we can calculate:
\[ y''(2) - 2y'(2) = [4(\ln 2 + 1)^2 + 2] - 2[4(\ln 2 + 1)] \]
\[ = 4(\ln 2 + 1)^2 + 2 - 8(\ln 2 + 1) \]

### Step 6: Simplify the expression

Let’s simplify:
\[ = 4(\ln 2 + 1)^2 - 8(\ln 2 + 1) + 2 \]

This is a quadratic in terms of \( \ln 2 + 1 \):
Let \( u = \ln 2 + 1 \):
\[ = 4u^2 - 8u + 2 \]

### Step 7: Factor or use the quadratic formula

We can use the quadratic formula:
\[ = 4(u^2 - 2u + \frac{1}{2}) = 4\left((u - 1)^2 - \frac{1}{2}\right) \]

### Final Result

Thus, the final answer is:
\[ y''(2) - 2y'(2) = 4\left((\ln 2 + 1 - 1)^2 - \frac{1}{2}\right) = 4\left((\ln 2)^2 - \frac{1}{2}\right) \]