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

Options

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

Answer

To differentiate the function \( y = x^{3e^{3x}} \) with respect to \( x \), we will use logarithmic differentiation. Here are the steps:

### 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: Apply the logarithmic property
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
Now, we differentiate both sides with respect to \( x \). Using implicit differentiation on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(3e^{3x} \ln x) \]

### Step 4: Differentiate the right side using the product rule
Let \( u = 3e^{3x} \) and \( v = \ln x \). Then, using the product rule \( \frac{d}{dx}(uv) = u'v + uv' \):
- \( u' = 9e^{3x} \) (since \( \frac{d}{dx}(3e^{3x}) = 3 \cdot 3e^{3x} \))
- \( v' = \frac{1}{x} \) (since \( \frac{d}{dx}(\ln x) = \frac{1}{x} \))

Now applying the product rule:
\[ \frac{d}{dx}(3e^{3x} \ln x) = 9e^{3x} \ln x + 3e^{3x} \cdot \frac{1}{x} \]

### Step 5: Substitute back into the equation
Substituting back into our differentiated equation:
\[ \frac{1}{y} \frac{dy}{dx} = 9e^{3x} \ln x + \frac{3e^{3x}}{x} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left( 9e^{3x} \ln x + \frac{3e^{3x}}{x} \right) \]

### Step 7: Substitute \( y \) back in
Recall that \( y = x^{3e^{3x}} \):
\[ \frac{dy}{dx} = x^{3e^{3x}} \left( 9e^{3x} \ln x + \frac{3e^{3x}}{x} \right) \]

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