If `y=2^(3^(x))`, then `y'` equals

Options

A. `3^(x).log3.log2`
B. `y.(.log2y).log3.log2`
C. `2^(3^(x)).3^(x).log6`
D. `2^(3^(x)).3^(x).log3.log2`

Answer

To find the derivative \( y' \) of the function \( y = 2^{3^x} \), we will use the chain rule and the properties of logarithms.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = 2^{3^x} \]

2. **Let \( g(x) = 3^x \)**:
We can rewrite \( y \) in terms of \( g(x) \):
\[ y = 2^{g(x)} \]

3. **Differentiate using the chain rule**:
The derivative of \( y = a^{u} \) is given by:
\[ \frac{dy}{dx} = a^u \ln(a) \cdot \frac{du}{dx} \]
Here, \( a = 2 \) and \( u = g(x) = 3^x \).

4. **Differentiate \( g(x) = 3^x \)**:
Using the formula for the derivative of an exponential function:
\[ \frac{dg}{dx} = 3^x \ln(3) \]

5. **Substituting back into the derivative of \( y \)**:
Now we can substitute \( g(x) \) and \( \frac{dg}{dx} \) back into our derivative formula:
\[ y' = 2^{g(x)} \ln(2) \cdot \frac{dg}{dx} \]
Substituting \( g(x) = 3^x \) and \( \frac{dg}{dx} = 3^x \ln(3) \):
\[ y' = 2^{3^x} \ln(2) \cdot (3^x \ln(3)) \]

6. **Final expression for \( y' \)**:
Thus, we can write:
\[ y' = 2^{3^x} \cdot 3^x \cdot \ln(2) \cdot \ln(3) \]

### Final Answer:
\[ y' = 2^{3^x} \cdot 3^x \cdot \ln(2) \cdot \ln(3) \]