Find `(d^(2)y)/(dx^(2))` when
`y=logx/x`.

Answer

To find the second derivative \(\frac{d^2y}{dx^2}\) when \(y = \frac{\log x}{x}\), we will follow these steps:

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

Given:
\[ y = \frac{\log x}{x} \]

We will use the quotient rule for differentiation, which states that if \(y = \frac{u}{v}\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Here, \(u = \log x\) and \(v = x\).

First, we find the derivatives:
- \(\frac{du}{dx} = \frac{1}{x}\)
- \(\frac{dv}{dx} = 1\)

Now applying the quotient rule:
\[ \frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} \]
\[ = \frac{1 - \log x}{x^2} \]

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

Now we need to differentiate \(\frac{dy}{dx}\) again:
\[ \frac{dy}{dx} = \frac{1 - \log x}{x^2} \]

Again, we will use the quotient rule:
Let \(u = 1 - \log x\) and \(v = x^2\).

Finding the derivatives:
- \(\frac{du}{dx} = -\frac{1}{x}\)
- \(\frac{dv}{dx} = 2x\)

Now applying the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
\[ = \frac{x^2 \left(-\frac{1}{x}\right) - (1 - \log x)(2x)}{(x^2)^2} \]
\[ = \frac{-x - 2x(1 - \log x)}{x^4} \]
\[ = \frac{-x - 2x + 2x \log x}{x^4} \]
\[ = \frac{-3x + 2x \log x}{x^4} \]

Now, we can simplify this:
\[ = \frac{x(-3 + 2 \log x)}{x^4} = \frac{-3 + 2 \log x}{x^3} \]

Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = \frac{-3 + 2 \log x}{x^3} \]

### Final Answer:
\[ \frac{d^2y}{dx^2} = \frac{-3 + 2 \log x}{x^3} \]

---