What is the d/(dt)(ae^tcos(2t)+be^tsin(2t)) ?

The d/(dt)(ae^tcos(2t)+be^tsin(2t)) is a(e^tcos(2t)-2e^tsin(2t))+b(e^tsin(2t)+cos(2t)*2e^t)

What is the first d/(dt)(ae^tcos(2t)+be^tsin(2t)) ?

The first d/(dt)(ae^tcos(2t)+be^tsin(2t)) is a(e^tcos(2t)-2e^tsin(2t))+b(e^tsin(2t)+cos(2t)*2e^t)

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\frac{d}{d t} (a e^{t} cos (2 t) + b e^{t} sin (2 t))

a (e^{t} cos (2 t) - 2 e^{t} sin (2 t)) + b (e^{t} sin (2 t) + cos (2 t) \cdot 2 e^{t})

Solution steps

\frac{d}{d t} (a e^{t} cos (2 t) + b e^{t} sin (2 t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (a e^{t} cos (2 t)) + \frac{d}{d t} (b e^{t} sin (2 t))

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

\frac{d}{d t} (b e^{t} sin (2 t)) = b (e^{t} sin (2 t) + cos (2 t) \cdot 2 e^{t})

= a (e^{t} cos (2 t) - 2 e^{t} sin (2 t)) + b (e^{t} sin (2 t) + cos (2 t) \cdot 2 e^{t})

x \to 0 lim ((1 + x)^{2} - 1)

x \to 1 lim (6 x^{2} + 1)

\int \frac{1}{4 + v ^{2}} d v

f (x) = \frac{1 - x ^{2}}{x}

What is the d/(dt)(ae^tcos(2t)+be^tsin(2t)) ?
The d/(dt)(ae^tcos(2t)+be^tsin(2t)) is a(e^tcos(2t)-2e^tsin(2t))+b(e^tsin(2t)+cos(2t)*2e^t)
What is the first d/(dt)(ae^tcos(2t)+be^tsin(2t)) ?
The first d/(dt)(ae^tcos(2t)+be^tsin(2t)) is a(e^tcos(2t)-2e^tsin(2t))+b(e^tsin(2t)+cos(2t)*2e^t)