What is (\partial)/(\partial t)(e^{-6t}cos(pix)) ?

The answer to (\partial)/(\partial t)(e^{-6t}cos(pix)) is -6e^{-6t}cos(pix)

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(\partial)/(\partial t)(e^{-6t}cos(pix))

\frac{\partial}{\partial t} (e^{- 6 t} cos (π x))

- 6 e^{- 6 t} cos (π x)

Solution steps

\frac{\partial}{\partial t} (e^{- 6 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^{- 6 t})

Apply the chain rule:

e^{- 6 t} \frac{\partial}{\partial t} (- 6 t)

= e^{- 6 t} \frac{\partial}{\partial t} (- 6 t)

\frac{\partial}{\partial t} (- 6 t) = - 6

= cos (π x) e^{- 6 t} (- 6)

Simplify

= - 6 e^{- 6 t} cos (π x)

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

\frac{9 ( 5 ^{x} )}{x ^{5}}

\int \frac{ln ( x ^{2} )}{x} d x

\int_{0}^{1} (arctan (2 x)) d x

f (x) = 7 x^{2} tan (x)

What is (\partial)/(\partial t)(e^{-6t}cos(pix)) ?
The answer to (\partial)/(\partial t)(e^{-6t}cos(pix)) is -6e^{-6t}cos(pix)