$\mathit{z}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{y}}$,$\mathit{x}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}\mathbf{ }$,$\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{t}$, find$\frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{t}}$.

The provided expressions are $z=x{e}^{-y}$,$x=\cosh t$, and $y=\cos t$.

The chain rule is a method for determining the derivative of composite functions, with the number of functions in the composition influencing how many differentiation steps are necessary.

Consider the equation given below.

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

The first given equation is differentiated with respect to x.

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial \left(x{e}^{-y}\right)}{\partial x} \\ =e^{-y}\frac{\partial x}{\partial x} \\ =e^{-y} \end{gathered}$$

The first given equation is differentiated with respect to y.

$$\begin{gathered} \frac{\partial z}{\partial y}=\frac{\partial \left(x{e}^{-y}\right)}{\partial y} \\ =x\frac{\partial e^{-y}}{\partial x} \\ =-x{e}^{-y} \\ =-z \end{gathered}$$

The second given equation is differentiated with respect to t.

$$\begin{gathered} \frac{dx}{dt}=\frac{d\left(\cos ht\right)}{dt} \\ =\sin ht \end{gathered}$$

The third given equation is differentiated with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d\left(\cos t\right)}{dt} \\ =\sin t \end{gathered}$$

Therefore, the value of $\frac{dz}{dt}$can be expressed as,

$\frac{dz}{dt}=\frac{dz}{dx}\frac{dx}{dt}+\frac{dz}{dy}\frac{dy}{dt}$

Substitute all the values in the above equation.

$$\begin{gathered} \frac{dz}{dt}=\left(e^{-t}\sin ht\right)+(-z)(-\sin t) \\ \frac{dz}{dt}=e^{-t}\sin ht+z\sin t \end{gathered}$$

Hence,$\frac{dz}{dt}=e^{-t}\sin ht+z\sin t$.