Given , $\{\begin{array}{l}2t+e^x=s-\text{cos}y-2\\ 2s-t=siny+x-1\end{array}$find ${\left(\frac{\partial s}{\partial t}\right)}_{\mathbf{y}}$at $\left(x,y,s,t\right)\mathbf{=}\left(0,\pi /2,-1,-2\right)$.

The given equations $2t+e^x=s-\cos y-2\left(1\right)$,$2s-t=\sin y+x-1\left(2\right)$ .

Differentiate the equations as:

$$\begin{gathered} 2dt+e^{x}dx=ds+\sin ydy\left(3\right) \\ 2ds-dt=\cos ydy+dx\left(4\right) \end{gathered}$$

Since, $y$is a constant, so$dy=0$ .

Substitute$dy=0$ in the equation (3) as:

$$\begin{gathered} 2dt+e^{x}dx=ds+\sin ydy \\ 2d{t}_y+e^{x}d{x}_y=d{s}_y+\sin y\left(0\right) \\ 2d{t}_y+e^{x}d{x}_y=d{s}_y \\ 2d{t}_y-d{s}_y=-e^{x}d{x}_y\left(5\right) \end{gathered}$$

Here, the subscript$y$ indicates that$y$ is a constant.

Substitute$dy=0$ in the equation (4) as:

$$\begin{gathered} 2ds-dt=\cos ydy+dx \\ 2d{s}_y-d{t}_y=\cos y\left(0\right)+d{x}_y \\ 2d{s}_y-d{t}_y=d{x}_y \\ d{x}_y=2d{s}_y-d{t}_y \end{gathered}$$

Substitute the value of$d{x}_y$ in the equation (5) as:

$$\begin{gathered} 2d{t}_y-d{s}_y=-e^{x}d{x}_y \\ 2d{t}_y-d{s}_y=-e^x\left(2d{s}_y-d{t}_y\right) \\ 2d{t}_y-d{s}_y=-e^{x}2d{s}_y+e^{x}d{t}_y \\ 2d{t}_y-e^{x}d{t}_y=-e^{x}2d{s}_y+d{s}_y \end{gathered}$$

Solving further as:

$$\begin{gathered} \left(2-e^x\right)d{t}_y=\left(-e^{x}2+1\right)d{s}_y \\ \frac{\left(2-e^x\right)}{\left(-e^{x}2+1\right)}d{t}_y=d{s}_y \end{gathered}$$

Now divide by$d{t}_y$ , then:

${\left(\frac{\partial s}{\partial t}\right)}_y=\frac{\left(2-e^x\right)}{\left(-e^{x}2+1\right)}\left(6\right)$

Substitute the values of $x$,$y$ ,$s$, and $t$in (6) equation as:

$$\begin{gathered} {\left(\frac{\partial s}{\partial t}\right)}_y=\frac{\left(2-e^x\right)}{\left(-e^{x}2+1\right)} \\ {\left(\frac{\partial s}{\partial t}\right)}_y=\frac{\left(2-e^0\right)}{\left(-e^{0}2+1\right)} \\ {\left(\frac{\partial s}{\partial t}\right)}_y=\frac{\left(2-1\right)}{\left(-2+1\right)} \\ {\left(\frac{\partial s}{\partial t}\right)}_y=-1 \end{gathered}$$

Therefore, the answer is${\left(\frac{\partial s}{\partial t}\right)}_y=-1$ .