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

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

Popular >

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

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

- e^{- t} cos (π x)

Solution steps

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

Apply the chain rule:

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

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

\frac{\partial}{\partial t} (- t) = - 1

= cos (π x) e^{- t} (- 1)

Simplify

= - e^{- t} cos (π x)

y = (\frac{6 - 5 x}{x ^{2} - 1})^{2}

\int \frac{1}{x ( x - 7 )} d x

\frac{d}{d y} (\frac{x ^{2} y ^{2} + 2 x y ^{3}}{( x + y ) ^{2}})

\int x sin (x) d x

\frac{d}{d x} (\frac{2 - sec ( x )}{tan ( x )})

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