Differentiate `x^(3e^(3x))` with respect to x.

Options

A. `3e^(3x)(3 log x + 1/x)`
B. `x^(3e^(3x)).3e^(3x)(3 log x - 1/x)`
C. `x^(3e^(3x))(3 log x + 1/x)`
D. `x^(3e^(3x)).3e^(3x)(3 log x + 1/x)`

Answer

To differentiate the function \( y = x^{3e^{3x}} \) with respect to \( x \), we can use logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides:
\[ \ln y = \ln(x^{3e^{3x}}) \]

### Step 2: Use the properties of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the equation:
\[ \ln y = 3e^{3x} \ln x \]

### Step 3: Differentiate both sides with respect to \( x \)
Now, we differentiate both sides with respect to \( x \). We will use the product rule on the right side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(3e^{3x} \ln x) \]

Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(3e^{3x}) \cdot \ln x + 3e^{3x} \cdot \frac{d}{dx}(\ln x) \]

### Step 4: Differentiate the right side
Now we differentiate \( 3e^{3x} \) and \( \ln x \):
- The derivative of \( 3e^{3x} \) is \( 9e^{3x} \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \).

Substituting these derivatives back into the equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = 9e^{3x} \ln x + 3e^{3x} \cdot \frac{1}{x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left(9e^{3x} \ln x + \frac{3e^{3x}}{x}\right) \]

### Step 6: Substitute back for \( y \)
Since \( y = x^{3e^{3x}} \), we substitute back:
\[ \frac{dy}{dx} = x^{3e^{3x}} \left(9e^{3x} \ln x + \frac{3e^{3x}}{x}\right) \]

### Final Result
Thus, the derivative of \( y = x^{3e^{3x}} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{3e^{3x}} \left(9e^{3x} \ln x + \frac{3e^{3x}}{x}\right) \]