`(d)/(dx)(3^(3^x))=`

Options

A. `2*3^(x)*3^(3^x)*llog3`
B. `3^(x)*(log3)^(2)`
C. `3^(3^x)*(log3)^(2)`
D. `3^(3^x)*3^(x)*(log3)^(2)`

Answer

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

### Step 1: Take the natural logarithm of both sides
We start with the equation:
\[ y = 3^{3^x} \]
Taking the natural logarithm of both sides gives:
\[ \ln y = \ln(3^{3^x}) \]

### Step 2: Apply the logarithmic identity
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can simplify the right-hand side:
\[ \ln y = 3^x \ln 3 \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. Remember that we need to use implicit differentiation on the left side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(3^x \ln 3) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \ln 3 \cdot \frac{d}{dx}(3^x) \]

### Step 4: Differentiate \( 3^x \)
Using the formula for the derivative of an exponential function, \( \frac{d}{dx}(a^x) = a^x \ln a \), we get:
\[ \frac{d}{dx}(3^x) = 3^x \ln 3 \]
So substituting this back into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \ln 3 \cdot (3^x \ln 3) \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \cdot (3^x \ln^2 3) \]
Substituting back \( y = 3^{3^x} \):
\[ \frac{dy}{dx} = 3^{3^x} \cdot (3^x \ln^2 3) \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = 3^{3^x} \cdot 3^x \ln^2 3 \]