1. Problem: Find the first and second derivatives $f'(x)$ and $f''(x)$ for the function $f(x) = 2^{2x+1}$. 2. Formula and rules: - For exponential functions with base $a$, $\frac{d}{dx} a^{g(x)} = a^{g(x)} \ln(a) g'(x)$. - The chain rule is used to differentiate composite functions. 3. First derivative $f'(x)$: - Given $f(x) = 2^{2x+1}$, let $g(x) = 2x+1$. - Then $g'(x) = 2$. - Using the formula: $$f'(x) = 2^{2x+1} \ln(2) \cdot 2 = 2^{2x+1} \cdot 2 \ln(2)$$ 4. Second derivative $f''(x)$: - Differentiate $f'(x)$: $$f''(x) = \frac{d}{dx} \left(2^{2x+1} \cdot 2 \ln(2)\right)$$ - Constants $2 \ln(2)$ can be factored out: $$f''(x) = 2 \ln(2) \cdot \frac{d}{dx} 2^{2x+1}$$ - Using the same rule as before: $$\frac{d}{dx} 2^{2x+1} = 2^{2x+1} \ln(2) \cdot 2$$ - Substitute back: $$f''(x) = 2 \ln(2) \cdot 2^{2x+1} \ln(2) \cdot 2 = 2^{2x+1} \cdot 4 \ln(2)^2$$ 5. Final answers: $$f'(x) = 2^{2x+1} \cdot 2 \ln(2)$$ $$f''(x) = 2^{2x+1} \cdot 4 \ln(2)^2$$ This completes the differentiation for the first function.