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

The d/(dt)(e^{-t}(2cos(t/2)-3sin(t/2))) is -7/2 e^{-t}cos(t/2)+2e^{-t}sin(t/2)

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

The first d/(dt)(e^{-t}(2cos(t/2)-3sin(t/2))) is -7/2 e^{-t}cos(t/2)+2e^{-t}sin(t/2)

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

- \frac{7}{2} e^{- t} cos (\frac{t}{2}) + 2 e^{- t} sin (\frac{t}{2})

Solution steps

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

Apply the Product Rule:

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

f = e^{- t}, g = 2 cos (\frac{t}{2}) - 3 sin (\frac{t}{2})

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

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

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

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

Simplify

(- e^{- t}) (2 cos (\frac{t}{2}) - 3 sin (\frac{t}{2})) + (- sin (\frac{t}{2}) - \frac{3}{2} cos (\frac{t}{2})) e^{- t} : - \frac{7}{2} e^{- t} cos (\frac{t}{2}) + 2 e^{- t} sin (\frac{t}{2})

= - \frac{7}{2} e^{- t} cos (\frac{t}{2}) + 2 e^{- t} sin (\frac{t}{2})

What is the d/(dt)(e^{-t}(2cos(t/2)-3sin(t/2))) ?
The d/(dt)(e^{-t}(2cos(t/2)-3sin(t/2))) is -7/2 e^{-t}cos(t/2)+2e^{-t}sin(t/2)
What is the first d/(dt)(e^{-t}(2cos(t/2)-3sin(t/2))) ?
The first d/(dt)(e^{-t}(2cos(t/2)-3sin(t/2))) is -7/2 e^{-t}cos(t/2)+2e^{-t}sin(t/2)