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Popular >

d/(dt)(({r}(t))(2sin(t)-e^{-t}))

Solution

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

Solution

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

Solution steps

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

Apply the Product Rule:

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

f = r (t), g = 2 sin (t) - e^{- t}

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

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

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

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Frequently Asked Questions (FAQ)

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