What is the derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} ?

The derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} is -e^{-t}sin(t)+e^{-2t}

What is the first derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} ?

The first derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} is -e^{-t}sin(t)+e^{-2t}

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derivative of

\frac{1}{2} e^{- t} cos (t) + \frac{1}{2} e^{- t} sin (t) - \frac{1}{2} e^{- 2 t}

- e^{- t} sin (t) + e^{- 2 t}

Solution steps

\frac{d}{d t} (\frac{1}{2} e^{- t} cos (t) + \frac{1}{2} e^{- t} sin (t) - \frac{1}{2} e^{- 2 t})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (\frac{1}{2} e^{- t} cos (t)) + \frac{d}{d t} (\frac{1}{2} e^{- t} sin (t)) - \frac{d}{d t} (\frac{1}{2} e^{- 2 t})

\frac{d}{d t} (\frac{1}{2} e^{- t} cos (t)) = \frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t))

\frac{d}{d t} (\frac{1}{2} e^{- t} sin (t)) = \frac{1}{2} (- e^{- t} sin (t) + e^{- t} cos (t))

\frac{d}{d t} (\frac{1}{2} e^{- 2 t}) = - e^{- 2 t}

= \frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t)) + \frac{1}{2} (- e^{- t} sin (t) + e^{- t} cos (t)) - (- e^{- 2 t})

Simplify

\frac{1}{2} (- e^{- t} cos (t) - e^{- t} sin (t)) + \frac{1}{2} (- e^{- t} sin (t) + e^{- t} cos (t)) - (- e^{- 2 t}) : - e^{- t} sin (t) + e^{- 2 t}

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

\frac{d}{d x} (cos^{2} (ln (x)))

y^{^{''}} + 8 y^{^{'}} + 6 y = 0

\frac{4 s + 3}{s ( s ^{2} + 6 s + 34 )}

f^{^{'}} (x) = x + 1

\frac{d}{d x} (- sin (2 x) \cdot 2 + cos (2 x) \cdot 2)

What is the derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} ?
The derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} is -e^{-t}sin(t)+e^{-2t}
What is the first derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} ?
The first derivative of 1/2 e^{-t}cos(t)+1/2 e^{-t}sin(t)-1/2 e^{-2t} is -e^{-t}sin(t)+e^{-2t}