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

Options

A. `(3-2 log x)/x^(3) `
B. `(2log x -3)/x^(3)`
C. `(2log x-3)/x^(4)`
D. `(2-3 log x)/(x^(4))`

Answer

To solve the problem of finding the second derivative of the function \( y = \frac{\log x}{x} \), we will follow these steps:

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

We will use the quotient rule for differentiation, which states that if \( y = \frac{u}{v} \), then:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

In our case, let \( u = \log x \) and \( v = x \).

- The derivative of \( u \) is \( \frac{du}{dx} = \frac{1}{x} \).
- The derivative of \( v \) is \( \frac{dv}{dx} = 1 \).

Now applying the quotient rule:

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

This simplifies to:

\[ \frac{dy}{dx} = \frac{1 - \log x}{x^2} \]

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

Now we need to differentiate \( \frac{dy}{dx} = \frac{1 - \log x}{x^2} \) again using the quotient rule.

Let \( u = 1 - \log x \) and \( v = x^2 \).

- The derivative of \( u \) is \( \frac{du}{dx} = -\frac{1}{x} \).
- The derivative of \( v \) is \( \frac{dv}{dx} = 2x \).

Now applying the quotient rule again:

\[ \frac{d^2y}{dx^2} = \frac{x^2 \cdot \left(-\frac{1}{x}\right) - (1 - \log x) \cdot 2x}{(x^2)^2} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = \frac{-x - 2x + 2x \log x}{x^4} \]

Combining the terms in the numerator gives:

\[ \frac{d^2y}{dx^2} = \frac{2x \log x - 3x}{x^4} \]

### Step 3: Simplify the expression

We can factor out \( x \) from the numerator:

\[ \frac{d^2y}{dx^2} = \frac{x(2 \log x - 3)}{x^4} \]

This simplifies to:

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

### Final Answer

Thus, the second derivative is:

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

---