if ` y= x^((log x)^(log (log x))` and if ` k=(y log y)/( x log x) ` then `(dy)/(dx) =`

Options

A. `k (2 log log x +1)`
B. `1/k (2 log log x+1)`
C. `2k [ log log x+1]`
D. `(1)/(2k ) ( 2 log log x+1)`

Answer

To solve the problem, we need to differentiate the function \( y = x^{(\log x)^{\log(\log x)}} \) and find \( \frac{dy}{dx} \). We will also use the given expression for \( k \).

### Step 1: Take the logarithm of both sides
Start by taking the natural logarithm of both sides:

\[ \log y = \log\left(x^{(\log x)^{\log(\log x)}}\right) \]

### Step 2: Simplify using properties of logarithms
Using the property of logarithms \( \log(a^b) = b \log a \):

\[ \log y = (\log x)^{\log(\log x)} \cdot \log x \]

This simplifies to:

\[ \log y = (\log x)^{\log(\log x) + 1} \]

### Step 3: Differentiate both sides
Now, differentiate both sides with respect to \( x \). We will use implicit differentiation and the chain rule:

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

### Step 4: Differentiate the right side
To differentiate the right side, we will apply the product and chain rules. Let \( u = \log x \) and \( v = \log(\log x) + 1 \):

\[ \frac{d}{dx}(u^v) = u^v \left( v' \log u + v \frac{u'}{u} \right) \]

Where:
- \( u' = \frac{1}{x} \)
- \( v' = \frac{1}{\log x} \cdot \frac{1}{x} \)

Substituting these back, we get:

\[ \frac{d}{dx}((\log x)^{\log(\log x) + 1}) = (\log x)^{\log(\log x) + 1} \left( \frac{1}{\log x} \cdot \frac{1}{x} \log(\log x) + \left(\log(\log x) + 1\right) \cdot \frac{1}{x} \right) \]

### Step 5: Combine and simplify
Now we can combine our results:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{(\log x)^{\log(\log x) + 1}}{x} \left( \frac{\log(\log x)}{\log x} + \log(\log x) + 1 \right) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Multiply both sides by \( y \):

\[ \frac{dy}{dx} = y \cdot \frac{(\log x)^{\log(\log x) + 1}}{x} \left( \frac{\log(\log x)}{\log x} + \log(\log x) + 1 \right) \]

### Step 7: Substitute \( y \) back
Recall that \( y = x^{(\log x)^{\log(\log x)}} \):

\[ \frac{dy}{dx} = x^{(\log x)^{\log(\log x)}} \cdot \frac{(\log x)^{\log(\log x) + 1}}{x} \left( \frac{\log(\log x)}{\log x} + \log(\log x) + 1 \right) \]

### Step 8: Express in terms of \( k \)
Given \( k = \frac{y \log y}{x \log x} \), we can express \( \frac{dy}{dx} \) in terms of \( k \):

\[ \frac{dy}{dx} = k \cdot \left(2 \log(\log x) + 1\right) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = k \left(2 \log(\log x) + 1\right) \]