If `f(x)=2^(sinx)`, then what is the derivative of f(x)?

Options

A. `2^(sinx)ln2`
B. `(sinx)2^(sinx-1)`
C. `(cosx)2^(sin x-1)`
D. None of the above

Explanation

To find the derivative of the function \( f(x) = 2^{\sin x} \), we will use the chain rule and the property of derivatives of exponential functions.

### Step-by-Step Solution:

1. **Identify the function**:
We have \( f(x) = 2^{\sin x} \).

2. **Use the derivative of an exponential function**:
The derivative of \( a^{g(x)} \) is given by:
\[ \frac{d}{dx}[a^{g(x)}] = a^{g(x)} \cdot \ln(a) \cdot g'(x) \]
Here, \( a = 2 \) and \( g(x) = \sin x \).

3. **Differentiate \( g(x) = \sin x \)**:
The derivative of \( g(x) = \sin x \) is:
\[ g'(x) = \cos x \]

4. **Apply the formula**:
Now, substituting \( a = 2 \) and \( g'(x) = \cos x \) into the derivative formula:
\[ f'(x) = 2^{\sin x} \cdot \ln(2) \cdot \cos x \]

5. **Final result**:
Therefore, the derivative of \( f(x) = 2^{\sin x} \) is:
\[ f'(x) = 2^{\sin x} \cdot \ln(2) \cdot \cos x \]