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

Options

A. ` (1)/(x) `
B. ` (-1)/(x^(2))`
C. ` 1`
D. `-1`

Answer

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

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Using the product rule for differentiation, where if \( u = x \) and \( v = \log x \), then:

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

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

- \( u = x \) implies \( \frac{du}{dx} = 1 \)
- \( v = \log x \) implies \( \frac{dv}{dx} = \frac{1}{x} \)

Now applying the product rule:

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

This simplifies to:

\[ \frac{dy}{dx} = 1 + \log x \]

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

Now we differentiate \( \frac{dy}{dx} = 1 + \log x \):

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

Since the derivative of a constant (1) is 0, we only need to differentiate \( \log x \):

\[ \frac{d^2y}{dx^2} = 0 + \frac{1}{x} \]

Thus, we have:

\[ \frac{d^2y}{dx^2} = \frac{1}{x} \]

### Final Answer

\[ \frac{d^2y}{dx^2} = \frac{1}{x} \]

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