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

Options

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

Answer

To solve the problem where \( y = \log(\log x) \) and we need to find \( \frac{d^2y}{dx^2} \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = \log(\log x) \]

We will use the chain rule to differentiate \( y \):
\[ \frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) \]

Now, we know that:
\[ \frac{d}{dx}(\log x) = \frac{1}{x} \]

So we substitute this into our equation:
\[ \frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now, we need to differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x \log x}\right) \]

Using the quotient rule, where \( u = 1 \) and \( v = x \log x \):
\[ \frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \):
\[ \frac{du}{dx} = 0 \]

Now we need to find \( \frac{dv}{dx} \):
\[ v = x \log x \implies \frac{dv}{dx} = \log x + 1 \]

Substituting back into the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(x \log x)(0) - (1)(\log x + 1)}{(x \log x)^2} \]
\[ = \frac{- (\log x + 1)}{(x \log x)^2} \]

### Final Result

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