What is the derivative of 6e^{-2x}cos(4x) ?

The derivative of 6e^{-2x}cos(4x) is 6(-2e^{-2x}cos(4x)-4e^{-2x}sin(4x))

What is the first derivative of 6e^{-2x}cos(4x) ?

The first derivative of 6e^{-2x}cos(4x) is 6(-2e^{-2x}cos(4x)-4e^{-2x}sin(4x))

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derivative of

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

6 (- 2 e^{- 2 x} cos (4 x) - 4 e^{- 2 x} sin (4 x))

Solution steps

\frac{d}{d x} (6 e^{- 2 x} cos (4 x))

Take the constant out:

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

= 6 \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)

= 6 (\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

= 6 ((- 2 e^{- 2 x}) cos (4 x) + (- sin (4 x) \cdot 4) e^{- 2 x})

Simplify

= 6 (- 2 e^{- 2 x} cos (4 x) - 4 e^{- 2 x} sin (4 x))

f (x) = 5 x^{2} - 2 x - 7

y^{^{'}} = 0.04 y - 2000, y (0) = 25000

e^{3 (t - 2)}

\frac{d}{d x} (\frac{2 x}{3 + x ^{2}})

\int x \cdot 1 - x^{2} d x

What is the derivative of 6e^{-2x}cos(4x) ?
The derivative of 6e^{-2x}cos(4x) is 6(-2e^{-2x}cos(4x)-4e^{-2x}sin(4x))
What is the first derivative of 6e^{-2x}cos(4x) ?
The first derivative of 6e^{-2x}cos(4x) is 6(-2e^{-2x}cos(4x)-4e^{-2x}sin(4x))