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 will differentiate the function \( y = \sqrt{x \log_e x} \) and evaluate \( \frac{dy}{dx} \) at \( x = e^4 \). Finally, we will calculate \( 4e^3k \).

### Step 1: Differentiate \( y \)
Given:
\[ y = \sqrt{x \log_e x} \]
We can rewrite this as:
\[ y = (x \log_e x)^{1/2} \]

Using the chain rule, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{2}(x \log_e x)^{-1/2} \cdot \frac{d}{dx}(x \log_e x) \]

### Step 2: Differentiate \( x \log_e x \)
To differentiate \( x \log_e x \), we use the product rule:
\[ \frac{d}{dx}(x \log_e x) = \log_e x + x \cdot \frac{d}{dx}(\log_e x) \]
The derivative of \( \log_e x \) is \( \frac{1}{x} \), so:
\[ \frac{d}{dx}(x \log_e x) = \log_e x + x \cdot \frac{1}{x} = \log_e x + 1 \]

### Step 3: Substitute back into the derivative
Now substituting back, we have:
\[ \frac{dy}{dx} = \frac{1}{2}(x \log_e x)^{-1/2} \cdot (\log_e x + 1) \]

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

### Step 5: Set \( k \)
From the above, we have:
\[ k = \frac{5}{4e^2} \]

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

### Step 7: Substitute \( e \)
Given \( e \approx 2.7 \):
\[ 4e^3k = 5 \cdot 2.7 = 13.5 \]

### Final Answer
Thus, the value of \( 4e^3k \) is:
\[ \boxed{13.5} \]