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

The derivative of y=-(e^{-x})/2 cos(2x) is 1/2 e^{-x}cos(2x)+e^{-x}sin(2x)

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

The first derivative of y=-(e^{-x})/2 cos(2x) is 1/2 e^{-x}cos(2x)+e^{-x}sin(2x)

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derivative of

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

\frac{1}{2} e^{- x} cos (2 x) + e^{- x} sin (2 x)

Solution steps

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

Simplify

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

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

Take the constant out:

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

= - \frac{1}{2} \frac{d}{d x} (e^{- x} cos (2 x))

Apply the Product Rule:

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

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

= - \frac{1}{2} (\frac{d}{d x} (e^{- x}) cos (2 x) + \frac{d}{d x} (cos (2 x)) e^{- x})

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

\frac{d}{d x} (cos (2 x)) = - sin (2 x) \cdot 2

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

Simplify

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

= \frac{1}{2} e^{- x} cos (2 x) + e^{- x} sin (2 x)

f (x) = x^{3} - 4 x^{2} - 1

\frac{d y}{d t} = - \frac{y}{t} - 8

\frac{d}{d x} (e^{4 x} cos (3 x))

\int sin (4 x) cos (4 x) d x

What is the derivative of y=-(e^{-x})/2 cos(2x) ?
The derivative of y=-(e^{-x})/2 cos(2x) is 1/2 e^{-x}cos(2x)+e^{-x}sin(2x)
What is the first derivative of y=-(e^{-x})/2 cos(2x) ?
The first derivative of y=-(e^{-x})/2 cos(2x) is 1/2 e^{-x}cos(2x)+e^{-x}sin(2x)