If ` y= x^(3)log x,then ( d^(2)y)/(dx^(2)) =`

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 \( \frac{dy}{dx} \)

Using the product rule, we differentiate \( y \):
\[ y = u \cdot v \quad \text{where } u = x^3 \text{ and } v = \log x \]
The product rule states that:
\[ \frac{dy}{dx} = u'v + uv' \]
Calculating \( u' \) and \( v' \):
\[ u' = \frac{d}{dx}(x^3) = 3x^2 \]
\[ v' = \frac{d}{dx}(\log x) = \frac{1}{x} \]
Now substituting back into the product rule:
\[ \frac{dy}{dx} = (3x^2)(\log x) + (x^3)\left(\frac{1}{x}\right) \]
\[ = 3x^2 \log x + x^2 \]
\[ = x^2(3 \log x + 1) \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} = x^2(3 \log x + 1) \) again using the product rule:
Let \( u = x^2 \) and \( v = 3 \log x + 1 \):
\[ \frac{d^2y}{dx^2} = u'v + uv' \]
Calculating \( u' \) and \( v' \):
\[ u' = \frac{d}{dx}(x^2) = 2x \]
For \( v' \):
\[ v' = \frac{d}{dx}(3 \log x + 1) = 3 \cdot \frac{1}{x} = \frac{3}{x} \]
Now substituting back into the product rule:
\[ \frac{d^2y}{dx^2} = (2x)(3 \log x + 1) + (x^2)\left(\frac{3}{x}\right) \]
\[ = 2x(3 \log x + 1) + 3x \]
Expanding this:
\[ = 6x \log x + 2x + 3x \]
\[ = 6x \log x + 5x \]

### Step 3: Factor out \( x \)

Factoring \( x \) from the expression:
\[ \frac{d^2y}{dx^2} = x(6 \log x + 5) \]

### Final Answer
Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = x(6 \log x + 5) \]

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