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

The d/(dt)(e^{-2t}(A*cos(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}(A*cos(4t)+B*,4t)) ?

The first d/(dt)(e^{-2t}(A*cos(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 \cdot cos (4 t) + B \cdot, 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}

5 x^{3} y^{2} x y

e v a l u a t e (- 2)^{4}

- \frac{3}{2} e^{- 2 (- 1)}

(e^{4})^{x} \cdot e^{x^{2}}

0.12 \cdot (8.2 \cdot 1 0^{- 2})

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