If `y=x^((logx)^log(logx))` then `(dy)/(dx)=`

Options

A. `(y)/(x)((In x^(x-1))"+2 In x In(In x))"`
B. `(y)/(x)(log x)^(log (log x))(2 log (log x )+1)`
C. `(y)/(" x In x")[(In x )^(2)+2In (In x)]`
D. `(y)/(x)(log y)/(log x)[2 log (log x)+1]`

Explanation

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

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

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

### Step 2: Use the power rule of logarithms
Using the power rule of logarithms, we can simplify the right-hand side:

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

### Step 3: Rewrite the equation
This simplifies to:

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

### Step 4: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). Using implicit differentiation on the left side and the product rule on the right side:

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

### Step 5: Differentiate the right-hand side
Using the chain rule and product rule, we differentiate:

Let \( u = \log x \) and \( v = \log(\log x) + 1 \). Then we have:

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

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

Substituting these into our equation gives:

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

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we can isolate \( \frac{dy}{dx} \):

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

### Step 7: Substitute back for \( y \)
Recall that \( y = x^{(\log x)^{\log(\log x)}} \). So we can substitute this back into our equation:

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

### Step 8: Simplify the expression
This simplifies to:

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

### Final Result
The final expression for \( \frac{dy}{dx} \) is:

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