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^(sinx^(-1))`
D. None of these

Answer

To find the derivative of the function \( f(x) = 2^{\sin x} \), we will follow these steps:

### Step 1: Identify the function and apply the chain rule
The function \( f(x) = 2^{\sin x} \) can be differentiated using the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by \( f'(g(x)) \cdot g'(x) \).

### Step 2: Differentiate the outer function
The outer function is \( 2^u \) where \( u = \sin x \). The derivative of \( 2^u \) with respect to \( u \) is \( 2^u \ln(2) \).

### Step 3: Differentiate the inner function
The inner function is \( u = \sin x \). The derivative of \( \sin x \) with respect to \( x \) is \( \cos x \).

### Step 4: Apply the chain rule
Now, we can apply the chain rule:
\[ f'(x) = \frac{d}{dx}(2^{\sin x}) = 2^{\sin x} \ln(2) \cdot \cos x \]

### Final Result
Thus, the derivative of \( f(x) = 2^{\sin x} \) is:
\[ f'(x) = 2^{\sin x} \ln(2) \cos x \]