Differentiate the following w.r.t. x:
`e^(2logx+3x)`

Answer

To differentiate the function \( y = e^{2 \log x + 3x} \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ y = e^{2 \log x + 3x} \]

### Step 2: Differentiate using the chain rule
Using the chain rule, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = e^{2 \log x + 3x} \cdot \frac{d}{dx}(2 \log x + 3x) \]

### Step 3: Differentiate the exponent
Now we need to differentiate the exponent \( 2 \log x + 3x \):
- The derivative of \( 2 \log x \) is \( \frac{2}{x} \) (using the derivative of \( \log x \)).
- The derivative of \( 3x \) is \( 3 \).

So, we have:
\[ \frac{d}{dx}(2 \log x + 3x) = \frac{2}{x} + 3 \]

### Step 4: Substitute back into the derivative
Now we substitute back into the derivative:
\[ \frac{dy}{dx} = e^{2 \log x + 3x} \left( \frac{2}{x} + 3 \right) \]

### Step 5: Final expression
Thus, the final expression for the derivative is:
\[ \frac{dy}{dx} = e^{2 \log x + 3x} \left( \frac{2}{x} + 3 \right) \]

### Step 6: Simplifying the expression (optional)
We can simplify \( e^{2 \log x} \):
\[ e^{2 \log x} = (e^{\log x})^2 = x^2 \]
So, the derivative can also be expressed as:
\[ \frac{dy}{dx} = x^2 e^{3x} \left( \frac{2}{x} + 3 \right) \]

### Final Answer
The final answer for the derivative is:
\[ \frac{dy}{dx} = x^2 e^{3x} \left( \frac{2}{x} + 3 \right) \]

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