Answer
$dz=-2e^{-2x} \cos(2 \pi t) dx -2 \pi e^{-2x}\sin(2 \pi t) dt$
Work Step by Step
Given the function $z=e^{-2x}cos(2 \pi t)$
The differential form can be evaluated as follows:
$dw=\dfrac{\partial w}{\partial x} dx + \dfrac{\partial w}{\partial y} dy + \dfrac{\partial w}{\partial z} dz$
We need to find the partial derivatives w.r.t. $t$ and $x$ as follows:
$f_x=-2e^{-2x} \cos(2 \pi t)\\f_t=-2 \pi e^{-2x}\sin(2 \pi t)$
Hence, we have
$dz=-2e^{-2x} \cos(2 \pi t) dx -2 \pi e^{-2x}\sin(2 \pi t) dt$
