Find the derivative of the following w.r.t. x :
`log(logx),xgt1`

Answer

To find the derivative of \( y = \log(\log x) \) with respect to \( x \) for \( x > 1 \), we will use the chain rule. Here’s the step-by-step solution:

### Step 1: Define the function
Let
\[ y = \log(\log x) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule. The chain rule states that if \( y = f(g(x)) \), then
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]

In our case, let \( u = \log x \). Then, we can rewrite \( y \) as:
\[ y = \log(u) \]

### Step 3: Differentiate \( y \) with respect to \( u \)
The derivative of \( y \) with respect to \( u \) is:
\[ \frac{dy}{du} = \frac{1}{u} \]

### Step 4: Differentiate \( u \) with respect to \( x \)
Now, we need to find \( \frac{du}{dx} \):
\[ u = \log x \implies \frac{du}{dx} = \frac{1}{x} \]

### Step 5: Apply the chain rule
Now, we can combine these derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{1}{x} \]

### Step 6: Substitute back for \( u \)
Since \( u = \log x \), we substitute back:
\[ \frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]

### Final Answer
Thus, the derivative of \( y = \log(\log x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{x \log x} \]