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

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

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

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

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derivative of

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

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

Solution steps

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

Simplify

e^{- 4 x} cos (- 4 x) : e^{- 4 x} cos (4 x)

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

\int (x \cdot 9^{2 x^{2} - 8}) d x

\frac{\partial}{\partial x} (1 + x ln (x y - 9))

\frac{2 s + 2}{s ^{2} + 2 s + 5}

\frac{\partial}{\partial y} (\frac{x}{1 + y ^{2}})

16 + t

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