Find the second derivatives of the following functions :
`x^(3)logx`

Answer

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

### Step 1: Find the First Derivative
We start by applying the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative is given by:

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

Here, let:
- \( u = x^3 \)
- \( v = \log x \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

1. \( \frac{du}{dx} = 3x^2 \)
2. \( \frac{dv}{dx} = \frac{1}{x} \)

Now, applying the product rule:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting the values:

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

This simplifies to:

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

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

We have:

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

We will differentiate each term separately:

1. The derivative of \( x^2 \) is \( 2x \).
2. For \( 3x^2 \log x \), we again use the product rule:
- Let \( u = 3x^2 \) and \( v = \log x \).
- Then \( \frac{du}{dx} = 6x \) and \( \frac{dv}{dx} = \frac{1}{x} \).

Using the product rule:

\[ \frac{d}{dx}(3x^2 \log x) = 3x^2 \cdot \frac{1}{x} + \log x \cdot 6x \]

This simplifies to:

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

Combining both derivatives:

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

This simplifies to:

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

### Final Answer
Thus, the second derivative of the function \( y = x^3 \log x \) is:

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