91Ó°ÊÓ

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.

$dz=\frac{∂z}{∂x}dx+\frac{∂z}{∂y}dy$

The first given equation is differentiated with respect to x.

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

The first given equation is differentiated with respect to y.

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

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