Answer
$dz=-2e^{-2x}cos(2 \pi t) dx -2\sin(2 \pi t)e^{-2x} dt$
Work Step by Step
The differential is defined as:
$dw=\frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy + \frac{\partial w}{\partial z} dz$
Given the function $z=e^{-2x}cos(2 \pi t)$ . Thus, we find the partial derivatives w.r.t $t$ and $x$:
$f_x=-2e^{-2x}cos(2 \pi t)$
$f_t=-2\sin(2 \pi t)e^{-2x}$
Thus, the differential:
$dz=-2e^{-2x}cos(2 \pi t) dx -2\sin(2 \pi t)e^{-2x} dt$
