What is the d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) ?

The d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) is 50e^{-t}sin(t)

What is the first d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) ?

The first d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) is 50e^{-t}sin(t)

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\frac{d}{d t} (25 - 25 e^{- t} cos (t) - 25 e^{- t} sin (t))

50 e^{- t} sin (t)

Solution steps

\frac{d}{d t} (25 - 25 e^{- t} cos (t) - 25 e^{- t} sin (t))

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (25) - \frac{d}{d t} (25 e^{- t} cos (t)) - \frac{d}{d t} (25 e^{- t} sin (t))

\frac{d}{d t} (25) = 0

\frac{d}{d t} (25 e^{- t} cos (t)) = 25 (- e^{- t} cos (t) - e^{- t} sin (t))

\frac{d}{d t} (25 e^{- t} sin (t)) = 25 (- e^{- t} sin (t) + e^{- t} cos (t))

= 0 - 25 (- e^{- t} cos (t) - e^{- t} sin (t)) - 25 (- e^{- t} sin (t) + e^{- t} cos (t))

Simplify

0 - 25 (- e^{- t} cos (t) - e^{- t} sin (t)) - 25 (- e^{- t} sin (t) + e^{- t} cos (t)) : 50 e^{- t} sin (t)

= 50 e^{- t} sin (t)

\frac{\partial}{\partial v} (e^{u (v) + v})

\int_{2}^{9} (x^{3} - π x^{2}) d x

f (x) = 8 (e^{x} + e^{x} x), at x = 0

x^{2} y^{^{'}} + 3 x y = \frac{sin ( 2 x )}{x}

\int \frac{x ^{2} + 25}{9 x ^{2}} d x

What is the d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) ?
The d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) is 50e^{-t}sin(t)
What is the first d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) ?
The first d/(dt)(25-25e^{-t}cos(t)-25e^{-t}sin(t)) is 50e^{-t}sin(t)