Find derivative of given functions w.r.t. the independent variable x.
`y=e^(x) ln x`

Options

A. `(e^(x))/(x) + e^(x) lnx`
B. `(e^(x))/(x)-e^(x) ln x`
C. `y=e^(x)lnx`
D. `(e^(x))/(x)-x`

Answer

To find the derivative of the function \( y = e^x \ln x \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product \( uv \) is given by:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, we can identify:
- \( u = e^x \)
- \( v = \ln x \)

Now, we will differentiate each function separately.

### Step 1: Differentiate \( u = e^x \)
The derivative of \( e^x \) with respect to \( x \) is:
\[ \frac{du}{dx} = e^x \]

### Step 2: Differentiate \( v = \ln x \)
The derivative of \( \ln x \) with respect to \( x \) is:
\[ \frac{dv}{dx} = \frac{1}{x} \]

### Step 3: Apply the product rule
Now, we apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we found:
\[ \frac{dy}{dx} = e^x \cdot \frac{1}{x} + \ln x \cdot e^x \]

### Step 4: Simplify the expression
We can factor out \( e^x \) from the expression:
\[ \frac{dy}{dx} = e^x \left( \frac{1}{x} + \ln x \right) \]

Thus, the derivative of the function \( y = e^x \ln x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = e^x \left( \frac{1}{x} + \ln x \right) \]

### Final Answer:
\[ \frac{dy}{dx} = e^x \left( \frac{1}{x} + \ln x \right) \]

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