What is the d/(dt)(e^{-2t}(Acos(4t)+B,4t)) ?

The d/(dt)(e^{-2t}(Acos(4t)+B,4t)) is (-2e^{-2t})(Acos(4t)+B,4t)+(-4Asin(4t)+(B,4*1))e^{-2t}

What is the first d/(dt)(e^{-2t}(Acos(4t)+B,4t)) ?

The first d/(dt)(e^{-2t}(Acos(4t)+B,4t)) is (-2e^{-2t})(Acos(4t)+B,4t)+(-4Asin(4t)+(B,4*1))e^{-2t}

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\frac{d}{d t} (e^{- 2 t} (A cos (4 t) + B, 4 t))

(- 2 e^{- 2 t}) (A cos (4 t) + B, 4 t) + (- 4 A sin (4 t) + (B, 4 \cdot 1)) e^{- 2 t}

Solution steps

\frac{d}{d t} (e^{- 2 t} (A cos (4 t) + B, 4 t))

Treat

A, B

as constants

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{- 2 t}, g = A cos (4 t) + B, 4 t

= \frac{d}{d t} (e^{- 2 t}) (A cos (4 t) + B, 4 t) + \frac{d}{d t} (A cos (4 t) + B, 4 t) e^{- 2 t}

\frac{d}{d t} (e^{- 2 t}) = - 2 e^{- 2 t}

\frac{d}{d t} (A cos (4 t) + B, 4 t) = - 4 A sin (4 t) + (B, 4 \cdot 1)

= (- 2 e^{- 2 t}) (A cos (4 t) + B, 4 t) + (- 4 A sin (4 t) + (B, 4 \cdot 1)) e^{- 2 t}

(x + 1) e^{- 1}

- 6 e^{- 6 (0)}

e^{- 0.1 (5)}

t^{3} e^{- x}

\frac{1}{13.5} \cdot 1 0^{- 6}

What is the d/(dt)(e^{-2t}(Acos(4t)+B,4t)) ?
The d/(dt)(e^{-2t}(Acos(4t)+B,4t)) is (-2e^{-2t})(Acos(4t)+B,4t)+(-4Asin(4t)+(B,4*1))e^{-2t}
What is the first d/(dt)(e^{-2t}(Acos(4t)+B,4t)) ?
The first d/(dt)(e^{-2t}(Acos(4t)+B,4t)) is (-2e^{-2t})(Acos(4t)+B,4t)+(-4Asin(4t)+(B,4*1))e^{-2t}