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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int x e^{2 x} sin (3 x) d x

\frac{2}{s ^{2} + 2 s + 1}

\frac{( z ^{n} )}{n !}

\int \frac{2 x}{3 x ^{2} - 1} d x

\frac{4 + x}{1 - 4 x}

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