What is the derivative of e^{-2x}cos(4x) ?

The derivative of e^{-2x}cos(4x) is -2e^{-2x}cos(4x)-4e^{-2x}sin(4x)

What is the first derivative of e^{-2x}cos(4x) ?

The first derivative of e^{-2x}cos(4x) is -2e^{-2x}cos(4x)-4e^{-2x}sin(4x)

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\frac{d}{d x} (e^{- 2 x} cos (4 x))

- 2 e^{- 2 x} cos (4 x) - 4 e^{- 2 x} sin (4 x)

Solution steps

\frac{d}{d x} (e^{- 2 x} cos (4 x))

Apply the Product Rule:

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

f = e^{- 2 x}, g = cos (4 x)

= \frac{d}{d x} (e^{- 2 x}) cos (4 x) + \frac{d}{d x} (cos (4 x)) e^{- 2 x}

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

\frac{d}{d x} (cos (4 x)) = - sin (4 x) \cdot 4

= (- 2 e^{- 2 x}) cos (4 x) + (- sin (4 x) \cdot 4) e^{- 2 x}

Simplify

= - 2 e^{- 2 x} cos (4 x) - 4 e^{- 2 x} sin (4 x)

\frac{d}{d x} (c e^{x})

f (x) = 5 x^{2} - x^{3}, at x = 1

2 sin (2 v) - 4 sin^{9} (2 v)

\frac{d}{d x} (x e^{- y})

\int_{- 2}^{- 1} \frac{2}{x ^{2}} d x

What is the derivative of e^{-2x}cos(4x) ?
The derivative of e^{-2x}cos(4x) is -2e^{-2x}cos(4x)-4e^{-2x}sin(4x)
What is the first derivative of e^{-2x}cos(4x) ?
The first derivative of e^{-2x}cos(4x) is -2e^{-2x}cos(4x)-4e^{-2x}sin(4x)