What is (\partial)/(\partial t)(e^{-4t}cos(pix)) ?

The answer to (\partial)/(\partial t)(e^{-4t}cos(pix)) is -4e^{-4t}cos(pix)

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(\partial)/(\partial t)(e^{-4t}cos(pix))

\frac{\partial}{\partial t} (e^{- 4 t} cos (π x))

- 4 e^{- 4 t} cos (π x)

Solution steps

\frac{\partial}{\partial t} (e^{- 4 t} cos (π x))

Treat

x

as a constant

Take the constant out:

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

= cos (π x) \frac{\partial}{\partial t} (e^{- 4 t})

Apply the chain rule:

e^{- 4 t} \frac{\partial}{\partial t} (- 4 t)

= e^{- 4 t} \frac{\partial}{\partial t} (- 4 t)

\frac{\partial}{\partial t} (- 4 t) = - 4

= cos (π x) e^{- 4 t} (- 4)

Simplify

= - 4 e^{- 4 t} cos (π x)

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 6 x}{x + 1})

x \to 1 lim \frac{( x ^{\frac{5}{2}} - 2 x ^{\frac{3}{2}} + x ^{\frac{1}{2}} )}{x - 1}

f (x) = (a - 5 x) (a + 5 x)

\int 4 sec^{6} (t) d t

2 y^{^{''}} + 19 y^{^{'}} - 10 y = 0

What is (\partial)/(\partial t)(e^{-4t}cos(pix)) ?
The answer to (\partial)/(\partial t)(e^{-4t}cos(pix)) is -4e^{-4t}cos(pix)