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

Options

A. ` y( 1+log (log x) )`
B. ` y(1+ logx )`
C. ` (y)/(x) (1+log (logx ) )`
D. ` (y)/(x) (1+log x)`

Answer

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

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the expression:
\[ \log y = \log((\log x)^{\log x}) \]

### Step 2: Apply the logarithmic power rule
Using the property of logarithms that states \( \log(a^b) = b \log a \), we can rewrite the right side:
\[ \log y = \log x \cdot \log(\log x) \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(\log x \cdot \log(\log x)) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log(\log x) + \log x \cdot \frac{d}{dx}(\log(\log x)) \]

### Step 4: Differentiate the right side
Now we differentiate each part:
- The derivative of \( \log x \) is \( \frac{1}{x} \).
- The derivative of \( \log(\log x) \) using the chain rule is \( \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \).

Putting it all together, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log(\log x) + \log x \cdot \frac{1}{x \log x} \]
The second term simplifies to:
\[ \frac{1}{x} \]

### Step 5: Combine the terms
Thus, we can combine the terms:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log(\log x) + \frac{1}{x} \]
Factoring out \( \frac{1}{x} \):
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} (1 + \log(\log x)) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot \frac{1}{x} (1 + \log(\log x)) \]

### Step 7: Substitute back for \( y \)
Since \( y = (\log x)^{\log x} \), we substitute back:
\[ \frac{dy}{dx} = (\log x)^{\log x} \cdot \frac{1}{x} (1 + \log(\log x)) \]

### Final Answer
Thus, the final result is:
\[ \frac{dy}{dx} = \frac{(\log x)^{\log x}}{x} (1 + \log(\log x)) \]