Let `y=sqrt(xlog_(e)x)`. If the value of `(dy)/(dx)` at `x=e^(4)` is k, then the value of `4e^(3)k` is (use e = 2.7 )

Answer

To solve the problem step by step, we need to find the derivative of the function \( y = \sqrt{x \log_e x} \) and evaluate it at \( x = e^4 \). Then, we will calculate \( 4e^3 k \) where \( k \) is the value of \( \frac{dy}{dx} \) at \( x = e^4 \).

### Step 1: Rewrite the function
We start with the function:
\[ y = \sqrt{x \log_e x} \]
This can be rewritten as:
\[ y = (x \log_e x)^{1/2} \]

### Step 2: Differentiate using the chain rule
Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{1}{2}(x \log_e x)^{-1/2} \cdot \frac{d}{dx}(x \log_e x) \]
Now we need to differentiate \( x \log_e x \).

### Step 3: Differentiate \( x \log_e x \)
Using the product rule:
\[ \frac{d}{dx}(x \log_e x) = \frac{d}{dx}(x) \cdot \log_e x + x \cdot \frac{d}{dx}(\log_e x) \]
Calculating each part:
\[ \frac{d}{dx}(x) = 1 \quad \text{and} \quad \frac{d}{dx}(\log_e x) = \frac{1}{x} \]
Thus,
\[ \frac{d}{dx}(x \log_e x) = \log_e x + 1 \]

### Step 4: Substitute back into the derivative
Now substituting back, we have:
\[ \frac{dy}{dx} = \frac{1}{2}(x \log_e x)^{-1/2} (\log_e x + 1) \]

### Step 5: Evaluate at \( x = e^4 \)
Now we evaluate \( \frac{dy}{dx} \) at \( x = e^4 \):
\[ \log_e(e^4) = 4 \]
Thus,
\[ x \log_e x = e^4 \cdot 4 = 4e^4 \]
Now substituting:
\[ \frac{dy}{dx} = \frac{1}{2}(4e^4)^{-1/2} (4 + 1) = \frac{1}{2}(2e^2)^{-1} \cdot 5 = \frac{5}{4e^2} \]
So we have:
\[ k = \frac{5}{4e^2} \]

### Step 6: Calculate \( 4e^3 k \)
Now we calculate:
\[ 4e^3 k = 4e^3 \cdot \frac{5}{4e^2} = 5e \]

### Step 7: Substitute \( e \)
Given \( e \approx 2.7 \):
\[ 5e \approx 5 \cdot 2.7 = 13.5 \]

Thus, the final answer is:
\[ \boxed{13.5} \]