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

Options

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

Answer

To find the derivative of \( e^{x \log x} \), we will use the chain rule and the product rule. Here’s a step-by-step solution:

### Step 1: Identify the function
We have the function \( y = e^{x \log x} \).

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we apply the chain rule:
\[ \frac{dy}{dx} = e^{x \log x} \cdot \frac{d}{dx}(x \log x) \]

### Step 3: Differentiate \( x \log x \) using the product rule
Now we need to differentiate \( x \log x \). We will use the product rule here, which states that if \( u = x \) and \( v = \log x \), then:
\[ \frac{d}{dx}(uv) = u'v + uv' \]
where \( u' = \frac{d}{dx}(x) = 1 \) and \( v' = \frac{d}{dx}(\log x) = \frac{1}{x} \).

So, applying the product rule:
\[ \frac{d}{dx}(x \log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1 \]

### Step 4: Substitute back into the derivative
Now we substitute \( \frac{d}{dx}(x \log x) \) back into our derivative:
\[ \frac{dy}{dx} = e^{x \log x} \cdot (\log x + 1) \]

### Step 5: Final expression
Thus, the derivative of \( e^{x \log x} \) is:
\[ \frac{dy}{dx} = e^{x \log x} (\log x + 1) \]

### Summary
The final answer is:
\[ \frac{dy}{dx} = e^{x \log x} (\log x + 1) \]

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