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

Answer

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

### Step 1: Find the First Derivative
We start by differentiating \( y \) with respect to \( x \).

1. **Differentiate \( y = \log(\log x) \)**:
Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) \]
We know that \( \frac{d}{dx}(\log x) = \frac{1}{x} \). Therefore,
\[ \frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]

### Step 2: Find the Second Derivative
Next, we differentiate \( \frac{dy}{dx} \) to find the second derivative.

2. **Differentiate \( \frac{dy}{dx} = \frac{1}{x \log x} \)**:
We will use the quotient rule here. The quotient rule states that if \( u = 1 \) and \( v = x \log x \), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, \( \frac{du}{dx} = 0 \) (since \( u = 1 \)) and we need to find \( \frac{dv}{dx} \).

- **Find \( v = x \log x \)**:
Using the product rule:
\[ \frac{dv}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1 \]
Therefore, we have:
\[ \frac{d}{dx}\left(\frac{1}{x \log x}\right) = \frac{(x \log x)(0) - (1)(\log x + 1)}{(x \log x)^2} \]
Simplifying this gives:
\[ \frac{d^2y}{dx^2} = \frac{- (\log x + 1)}{(x \log x)^2} \]

### Final Result
Thus, the second derivative of \( y = \log(\log x) \) is:
\[ \frac{d^2y}{dx^2} = \frac{- (\log x + 1)}{(x \log x)^2} \]

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