What is the derivative of e^{-2x}(cos(3x+sin(3x))) ?

The derivative of e^{-2x}(cos(3x+sin(3x))) is e^{-2x}cos(3x)-5e^{-2x}sin(3x)

What is the first derivative of e^{-2x}(cos(3x+sin(3x))) ?

The first derivative of e^{-2x}(cos(3x+sin(3x))) is e^{-2x}cos(3x)-5e^{-2x}sin(3x)

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\frac{d}{d x} (e^{- 2 x} (cos (3 x) + sin (3 x)))

e^{- 2 x} cos (3 x) - 5 e^{- 2 x} sin (3 x)

Solution steps

\frac{d}{d x} (e^{- 2 x} (cos (3 x) + sin (3 x)))

Apply the Product Rule:

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

f = e^{- 2 x}, g = cos (3 x) + sin (3 x)

= \frac{d}{d x} (e^{- 2 x}) (cos (3 x) + sin (3 x)) + \frac{d}{d x} (cos (3 x) + sin (3 x)) e^{- 2 x}

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

\frac{d}{d x} (cos (3 x) + sin (3 x)) = - sin (3 x) \cdot 3 + cos (3 x) \cdot 3

= (- 2 e^{- 2 x}) (cos (3 x) + sin (3 x)) + (- sin (3 x) \cdot 3 + cos (3 x) \cdot 3) e^{- 2 x}

Simplify

(- 2 e^{- 2 x}) (cos (3 x) + sin (3 x)) + (- sin (3 x) \cdot 3 + cos (3 x) \cdot 3) e^{- 2 x} : e^{- 2 x} cos (3 x) - 5 e^{- 2 x} sin (3 x)

= e^{- 2 x} cos (3 x) - 5 e^{- 2 x} sin (3 x)

\int \frac{4400 q}{4 q ^{2} + 320} d q

y^{^{'}} = - 0.5 y^{1.5}

\frac{\partial}{\partial y} (x arcsin (\frac{y}{x}))

f (x) = (x + 5) (9 x + 5)

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

What is the derivative of e^{-2x}(cos(3x+sin(3x))) ?
The derivative of e^{-2x}(cos(3x+sin(3x))) is e^{-2x}cos(3x)-5e^{-2x}sin(3x)
What is the first derivative of e^{-2x}(cos(3x+sin(3x))) ?
The first derivative of e^{-2x}(cos(3x+sin(3x))) is e^{-2x}cos(3x)-5e^{-2x}sin(3x)