What is the d/(dt)(1/2 e^{-4t}cos(8t)) ?

The d/(dt)(1/2 e^{-4t}cos(8t)) is -2e^{-4t}(cos(8t)+2sin(8t))

What is the first d/(dt)(1/2 e^{-4t}cos(8t)) ?

The first d/(dt)(1/2 e^{-4t}cos(8t)) is -2e^{-4t}(cos(8t)+2sin(8t))

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\frac{d}{d t} (\frac{1}{2} e^{- 4 t} cos (8 t))

- 2 e^{- 4 t} (cos (8 t) + 2 sin (8 t))

Solution steps

\frac{d}{d t} (\frac{1}{2} e^{- 4 t} cos (8 t))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d t} (e^{- 4 t} cos (8 t))

Apply the Product Rule:

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

f = e^{- 4 t}, g = cos (8 t)

= \frac{1}{2} (\frac{d}{d t} (e^{- 4 t}) cos (8 t) + \frac{d}{d t} (cos (8 t)) e^{- 4 t})

\frac{d}{d t} (e^{- 4 t}) = - 4 e^{- 4 t}

\frac{d}{d t} (cos (8 t)) = - sin (8 t) \cdot 8

= \frac{1}{2} ((- 4 e^{- 4 t}) cos (8 t) + (- sin (8 t) \cdot 8) e^{- 4 t})

Simplify

\frac{1}{2} ((- 4 e^{- 4 t}) cos (8 t) + (- sin (8 t) \cdot 8) e^{- 4 t}) : - 2 e^{- 4 t} (cos (8 t) + 2 sin (8 t))

= - 2 e^{- 4 t} (cos (8 t) + 2 sin (8 t))

\int_{0}^{1} π (x^{4} - x) d x

\frac{\partial}{\partial x} (e^{e^{y} x + y})

f (x) = 4 x^{2} + 6 x + 3

\int e^{- (x)^{2}} d x

\int 4 x sin (5 x) d x

What is the d/(dt)(1/2 e^{-4t}cos(8t)) ?
The d/(dt)(1/2 e^{-4t}cos(8t)) is -2e^{-4t}(cos(8t)+2sin(8t))
What is the first d/(dt)(1/2 e^{-4t}cos(8t)) ?
The first d/(dt)(1/2 e^{-4t}cos(8t)) is -2e^{-4t}(cos(8t)+2sin(8t))