Find the 2nd derivative if `x^3 log x` with respect to x.

Answer

To find the second derivative of the function \( y = x^3 \log x \) with respect to \( x \), we will follow these steps:

### Step 1: Find the First Derivative
We start by applying the product rule to differentiate \( y = x^3 \log x \). The product rule states that if \( u \) and \( v \) are functions of \( x \), then:

\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, let:
- \( u = x^3 \) and \( v = \log x \)

Now, we find the derivatives:
- \( \frac{du}{dx} = 3x^2 \)
- \( \frac{dv}{dx} = \frac{1}{x} \)

Applying the product rule, we have:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x^3 \cdot \frac{1}{x} + \log x \cdot 3x^2 \]

This simplifies to:

\[ \frac{dy}{dx} = x^2 + 3x^2 \log x \]

Factoring out \( x^2 \):

\[ \frac{dy}{dx} = x^2 (3 \log x + 1) \]

### Step 2: Find the Second Derivative
Now we differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \).

Using the product rule again on \( \frac{dy}{dx} = x^2 (3 \log x + 1) \):

Let:
- \( u = x^2 \) and \( v = 3 \log x + 1 \)

Now, we find the derivatives:
- \( \frac{du}{dx} = 2x \)
- \( \frac{dv}{dx} = \frac{3}{x} \)

Applying the product rule:

\[ \frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx} = x^2 \cdot \frac{3}{x} + (3 \log x + 1) \cdot 2x \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = 3x + 2x(3 \log x + 1) \]

Distributing \( 2x \):

\[ \frac{d^2y}{dx^2} = 3x + 6x \log x + 2x \]

Combining like terms:

\[ \frac{d^2y}{dx^2} = 5x + 6x \log x \]

### Final Result
Thus, the second derivative of \( y = x^3 \log x \) with respect to \( x \) is:

\[ \frac{d^2y}{dx^2} = x(6 \log x + 5) \]