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

Answer

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

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

Using 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 \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = \frac{1}{x} \)
- \( \frac{dv}{dx} = 1 \)

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

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

- We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = -\frac{1}{x} \)
- \( \frac{dv}{dx} = 2x \)

Now applying the quotient rule again:

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

Substituting in the values:

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

Simplifying the numerator:

\[ = \frac{-x - 2x(1 - \log x)}{x^4} \]

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

\[ = \frac{-3x + 2x \log x}{x^4} \]

Factoring out \( x \):

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

### Final Result

Thus, the second derivative is:

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

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