If `f(x)=2^(x)`, then what is f''(x) equal to ?

Options

A. `2^(x)(ln 2)^(2)`
B. `x(x-1)2^(x-2)`
C. `2^(x+1)(ln2)`
D. `2^(x)(log_(10)2)^(2)`

Explanation

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

### Step 1: Find the first derivative \( f'(x) \)

We start with the function:
\[ f(x) = 2^x \]

Using the formula for the derivative of an exponential function, where \( \frac{d}{dx}(a^x) = a^x \ln(a) \), we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2^x) = 2^x \ln(2) \]

### Step 2: Find the second derivative \( f''(x) \)

Now, we differentiate \( f'(x) \) to find the second derivative:
\[ f'(x) = 2^x \ln(2) \]

Using the product rule, where \( \frac{d}{dx}(u \cdot v) = u'v + uv' \), we differentiate \( f'(x) \):
- Let \( u = 2^x \) and \( v = \ln(2) \).
- The derivative of \( u = 2^x \) is \( u' = 2^x \ln(2) \).
- The derivative of \( v = \ln(2) \) is \( v' = 0 \) (since it is a constant).

Applying the product rule:
\[ f''(x) = u'v + uv' = (2^x \ln(2)) \ln(2) + (2^x)(0) \]
\[ f''(x) = 2^x \ln(2) \ln(2) = 2^x (\ln(2))^2 \]

### Final Result

Thus, the second derivative \( f''(x) \) is:
\[ f''(x) = 2^x (\ln(2))^2 \]

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