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Differentiate the following functions s=4e^3t-e^-2.5 w.r.t.t

Sagot :

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Answer:

[tex]\displaystyle \frac{ds}{dt} = 12e^{3t}[/tex]

General Formulas and Concepts:

Algebra I

  • Functions
  • Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

eˣ Derivative:                                                                                                         [tex]\displaystyle \frac{d}{dx} [e^u]=e^u \cdot u'[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle s = 4e^{3t} - e^{-2.5}[/tex]

Step 2: Differentiate

  1. eˣ Derivative:                                                                                                 [tex]\displaystyle \frac{ds}{dt} = 4e^{3t} \cdot \frac{d}{dt}[3t] - \frac{d}{dt}[e^{-2.5}][/tex]
  2. Basic Power Rule:                                                                                         [tex]\displaystyle \frac{ds}{dt} = 4e^{3t} \cdot 3t^{1 - 1} - 0[/tex]
  3. Simplify:                                                                                                         [tex]\displaystyle \frac{ds}{dt} = 12e^{3t}[/tex]

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

Unit: Derivatives

Book: College Calculus 10e

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