Answer:
\(\displaystyle F'(t) = 2e^{8t} \bigg( \cos (2x) + 4 \sin (2x) \bigg)\)
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: \(\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)\)
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: \(\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)\)
Derivative Rule [Chain Rule]: \(\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)\)
Step-by-step explanation:
Step 1: Define
Identify
\(\displaystyle F(t) = e^{8t} \sin 2t\)
Step 2: Differentiate
- Derivative Rule [Product Rule]: \(\displaystyle F'(t) = (e^{8t})' \sin 2t + e^{8t}(\sin 2t)'\)
- Exponential Differentiation [Derivative Rule - Chain Rule]: \(\displaystyle F'(t) = e^{8t}(8t)' \sin 2t + e^{8t}(\sin 2t)'\)
- Trigonometric Differentiation [Derivative Rule - Chain Rule]: \(\displaystyle F'(t) = e^{8t}(8t)' \sin 2t + e^{8t} \cos 2t (2t)'\)
- Basic Power Rule [Derivative Property - Multiplied Constant]: \(\displaystyle F'(t) = 8e^{8t} \sin 2t + 2e^{8t} \cos 2t\)
- Factor: \(\displaystyle F'(t) = 2e^{8t} \bigg( \cos (2x) + 4 \sin (2x) \bigg)\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
