Question

Derivative of F(t) = e8t sin 2t

Answer 1

`\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 (d)/(dx) [cf(x)] = c \cdot f'(x)`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
`\displaystyle (d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)`

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Explanation:

Step 1: Define

Identify

`\displaystyle F(t) = e^(8t) \sin 2t`

Step 2: Differentiate

1. Derivative Rule [Product Rule]:
   `\displaystyle F'(t) = (e^(8t))' \sin 2t + e^(8t)(\sin 2t)'`
2. Exponential Differentiation [Derivative Rule - Chain Rule]:
   `\displaystyle F'(t) = e^(8t)(8t)' \sin 2t + e^(8t)(\sin 2t)'`
3. Trigonometric Differentiation [Derivative Rule - Chain Rule]:
   `\displaystyle F'(t) = e^(8t)(8t)' \sin 2t + e^(8t) \cos 2t (2t)'`
4. Basic Power Rule [Derivative Property - Multiplied Constant]:
   `\displaystyle F'(t) = 8e^(8t) \sin 2t + 2e^(8t) \cos 2t`
5. Factor:
   `\displaystyle F'(t) = 2e^(8t) \bigg( \cos (2x) + 4 \sin (2x) \bigg)`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation