Derivative of `e^(x log x )` is :

Options

A. x log x
B. `x^x`
C. `e^(x log x )(1+log x)`
D. `x(x)^(x-1)`

Answer

To find the derivative of the function \( e^{x \log x} \), we can follow these steps:

### Step 1: Identify the function
We have the function:
\[ y = e^{x \log x} \]

### Step 2: Use the chain rule
To differentiate \( y \), we will use the chain rule. The derivative of \( e^u \) is \( e^u \cdot \frac{du}{dx} \), where \( u = x \log x \).

### Step 3: Differentiate \( u = x \log x \)
Now we need to find the derivative of \( u \):
\[ u = x \log x \]
Using the product rule, where \( f(x) = x \) and \( g(x) = \log x \):
\[ \frac{du}{dx} = f'(x)g(x) + f(x)g'(x) \]
Here, \( f'(x) = 1 \) and \( g'(x) = \frac{1}{x} \). Thus:
\[ \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1 \]

### Step 4: Apply the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = e^{x \log x} \cdot \frac{du}{dx} = e^{x \log x} \cdot (\log x + 1) \]

### Step 5: Simplify the expression
Since \( e^{x \log x} = x^x \) (because \( e^{\log a} = a \)), we can express the derivative as:
\[ \frac{dy}{dx} = x^x \cdot (\log x + 1) \]

### Final Result
Thus, the derivative of \( e^{x \log x} \) is:
\[ \frac{dy}{dx} = e^{x \log x} \cdot (1 + \log x) \]