What is the d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) ?

The d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) is e^{-2t}cos(3t)-5e^{-2t}sin(3t)

What is the first d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) ?

The first d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) is e^{-2t}cos(3t)-5e^{-2t}sin(3t)

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\frac{d}{d t} (e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t))

e^{- 2 t} cos (3 t) - 5 e^{- 2 t} sin (3 t)

Solution steps

\frac{d}{d t} (e^{- 2 t} cos (3 t) + e^{- 2 t} sin (3 t))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (e^{- 2 t} cos (3 t)) + \frac{d}{d t} (e^{- 2 t} sin (3 t))

\frac{d}{d t} (e^{- 2 t} cos (3 t)) = - 2 e^{- 2 t} cos (3 t) - 3 e^{- 2 t} sin (3 t)

\frac{d}{d t} (e^{- 2 t} sin (3 t)) = - 2 e^{- 2 t} sin (3 t) + 3 e^{- 2 t} cos (3 t)

= - 2 e^{- 2 t} cos (3 t) - 3 e^{- 2 t} sin (3 t) - 2 e^{- 2 t} sin (3 t) + 3 e^{- 2 t} cos (3 t)

Simplify

= e^{- 2 t} cos (3 t) - 5 e^{- 2 t} sin (3 t)

\frac{d}{d x} (- 2 x^{5} - 3 x^{3} + 1)

x \to 2 - lim (∣ x ∣ - x)

y^{^{''}} + \frac{2 y ^{^{'}}}{x} - \frac{6 y}{x ^{2}} = 0

x \to 1 lim (1 - cos (π) x)

\int 4 sin^{4} (x) d x

What is the d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) ?
The d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) is e^{-2t}cos(3t)-5e^{-2t}sin(3t)
What is the first d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) ?
The first d/(dt)(e^{-2t}cos(3t)+e^{-2t}sin(3t)) is e^{-2t}cos(3t)-5e^{-2t}sin(3t)