What is (\partial)/(\partial t)(e^{-2t}cos(2x)) ?

The answer to (\partial)/(\partial t)(e^{-2t}cos(2x)) is -2e^{-2t}cos(2x)

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(\partial)/(\partial t)(e^{-2t}cos(2x))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

e^{- 2 t} \frac{\partial}{\partial t} (- 2 t)

= e^{- 2 t} \frac{\partial}{\partial t} (- 2 t)

\frac{\partial}{\partial t} (- 2 t) = - 2

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

Simplify

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

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

\int (4 x^{2} + \frac{1}{( 4 x ) ^{2}}) d x

y^{^{'}} = 9 x 7 y + 8, y (0) = 3

\frac{d}{d x} (\frac{x \cdot cos ( x ) + π}{π - x})

\frac{t}{3}

What is (\partial)/(\partial t)(e^{-2t}cos(2x)) ?
The answer to (\partial)/(\partial t)(e^{-2t}cos(2x)) is -2e^{-2t}cos(2x)