Find the second derivatives of the following functions :
`x^(x)`

Answer

To find the second derivative of the function \( y = x^x \), we will follow these steps:

### Step 1: Rewrite the Function
Let \( y = x^x \).

### Step 2: Take the Natural Logarithm
Taking the natural logarithm on both sides:
\[ \ln y = \ln(x^x) \]
Using the property of logarithms, this simplifies to:
\[ \ln y = x \ln x \]

### Step 3: Differentiate Both Sides
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y(\ln x + 1) \]
Substituting back \( y = x^x \):
\[ \frac{dy}{dx} = x^x (\ln x + 1) \]

### Step 5: Differentiate Again to Find the Second Derivative
Now we differentiate \( \frac{dy}{dx} \) again to find the second derivative \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^x (\ln x + 1)) \]
Using the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^x) (\ln x + 1) + x^x \frac{d}{dx}(\ln x + 1) \]

### Step 6: Differentiate \( x^x \) Again
We already found \( \frac{dy}{dx} = x^x (\ln x + 1) \), so we can differentiate \( x^x \):
\[ \frac{d}{dx}(x^x) = x^x (\ln x + 1) \]
Thus,
\[ \frac{d^2y}{dx^2} = x^x (\ln x + 1)(\ln x + 1) + x^x \cdot \frac{1}{x} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = x^x (\ln x + 1)^2 + x^{x-1} \]

### Final Result
Therefore, the second derivative of \( y = x^x \) is:
\[ \frac{d^2y}{dx^2} = x^x \left( (\ln x + 1)^2 + \frac{1}{x} \right) \]