What is the (e^{-t}sin(t))^' ?

The (e^{-t}sin(t))^' is -e^{-t}sin(t)+e^{-t}cos(t)

What is the first (e^{-t}sin(t))^' ?

The first (e^{-t}sin(t))^' is -e^{-t}sin(t)+e^{-t}cos(t)

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(e^{- t} sin (t))^{^{'}}

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

Solution steps

(e^{- t} sin (t))^{'}

Apply the Product Rule:

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

f = e^{- t}, g = sin (t)

= (e^{- t})^{^{'}} sin (t) + (sin (t))^{^{'}} e^{- t}

(e^{- t})^{'} = - e^{- t}

(sin (t))^{'} = cos (t)

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

Simplify

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

ln (\frac{x}{x ^{2} + 4})

\frac{\partial}{\partial y} (2 (x + y))

\frac{\partial}{\partial x} (\frac{2 x - y}{x + 3 y})

x \to 2 - lim (x + c)

\int x - \frac{2}{x} d x

What is the (e^{-t}sin(t))^' ?
The (e^{-t}sin(t))^' is -e^{-t}sin(t)+e^{-t}cos(t)
What is the first (e^{-t}sin(t))^' ?
The first (e^{-t}sin(t))^' is -e^{-t}sin(t)+e^{-t}cos(t)