If `y=xlnx+"xe"^(x)`. then what is the value of `(dy)/(dx)` at `x=1` ?

Options

A. `1+e`
B. `1-e`
C. `1+2e`
D. None of these

Answer

To find the value of \(\frac{dy}{dx}\) at \(x=1\) for the function \(y = x \ln x + x e^x\), we will follow these steps:

### Step 1: Differentiate the function
We need to differentiate \(y\) with respect to \(x\). The function consists of two parts: \(x \ln x\) and \(x e^x\). We will use the product rule for differentiation, which states that if \(y = u \cdot v\), then \(\frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}\).

Let:
- \(u_1 = x\) and \(v_1 = \ln x\) for the first term \(x \ln x\)
- \(u_2 = x\) and \(v_2 = e^x\) for the second term \(x e^x\)

#### Differentiate \(x \ln x\):
Using the product rule:
\[ \frac{d}{dx}(x \ln x) = x \cdot \frac{d}{dx}(\ln x) + \ln x \cdot \frac{d}{dx}(x) \]
\[ = x \cdot \frac{1}{x} + \ln x \cdot 1 = 1 + \ln x \]

#### Differentiate \(x e^x\):
Using the product rule:
\[ \frac{d}{dx}(x e^x) = x \cdot \frac{d}{dx}(e^x) + e^x \cdot \frac{d}{dx}(x) \]
\[ = x e^x + e^x \]

### Step 2: Combine the derivatives
Now we combine the derivatives of both parts:
\[ \frac{dy}{dx} = (1 + \ln x) + (x e^x + e^x) \]
\[ = 1 + \ln x + e^x + x e^x \]

### Step 3: Substitute \(x = 1\)
Now we substitute \(x = 1\) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=1} = 1 + \ln(1) + e^1 + 1 \cdot e^1 \]
Since \(\ln(1) = 0\) and \(e^1 = e\):
\[ = 1 + 0 + e + e = 1 + 2e \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = 1\) is:
\[ \frac{dy}{dx} \bigg|_{x=1} = 1 + 2e \]