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Popular >

d/(dt)(e^{-t}cos(2t))

Solution

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

Solution

- e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

= - e^{- t} cos (2 t) - 2 e^{- t} sin (2 t)

Popular Examples

x^{''}+16pi^2x=0

x^{^{''}} + 16 π^{2} x = 0

integral of (2x+5)^4

\int (2 x + 5)^{4} d x

derivative of y= 4/(3x^2)derivative of

y = \frac{4}{3 x ^{2}}

derivative of 4/((x^2-2x)^2)derivative of

\frac{4}{( x ^{2} - 2 x ) ^{2}}

f(x)=-(e^{1/x})/(x^2)

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

Frequently Asked Questions (FAQ)

What is the d/(dt)(e^{-t}cos(2t)) ?
The d/(dt)(e^{-t}cos(2t)) is -e^{-t}cos(2t)-2e^{-t}sin(2t)
What is the first d/(dt)(e^{-t}cos(2t)) ?
The first d/(dt)(e^{-t}cos(2t)) is -e^{-t}cos(2t)-2e^{-t}sin(2t)