91Ó°ÊÓ

The given equations $w=f\left(x,s,t\right)$, $s=2x+y\left(1\right)$, $t=2x-y\left(2\right)$.

Differentiate the function as:

$dw=\frac{∂f}{∂x}dx+\frac{∂f}{∂s}ds+\frac{∂f}{∂t}dt\left(3\right)$

Also, differentiate equation (1) and (2) as:

$$\begin{gathered} ds=2dx+dy \\ dt=2dx-dy \end{gathered}$$

Substitute the values of $ds$and$dt$ in (3) then:

$$\begin{gathered} dw=\frac{∂f}{∂x}dx+\frac{∂f}{∂s}ds+\frac{∂f}{∂t}dt \\ dw=\frac{∂f}{∂x}dx+\frac{∂f}{∂s}\left(2dx+dy\right)+\frac{∂f}{∂t}\left(2dx-dy\right) \\ dw=\frac{∂f}{∂x}dx+2\frac{∂f}{∂s}dx+\frac{∂f}{∂s}dy+2\frac{∂f}{∂t}dx-\frac{∂f}{∂t}dy \\ dw=\left(\frac{∂f}{∂x}+2\frac{∂f}{∂s}+2\frac{∂f}{∂t}\right)dx+\left(\frac{∂f}{∂s}-\frac{∂f}{∂t}\right)dy\left(4\right) \end{gathered}$$

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

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

$$\begin{gathered} dw=\left(\frac{∂f}{∂x}+2\frac{∂f}{∂s}+2\frac{∂f}{∂t}\right)dx+\left(\frac{∂f}{∂s}-\frac{∂f}{∂t}\right)dy \\ dw=\left(\frac{∂f}{∂x}+2\frac{∂f}{∂s}+2\frac{∂f}{∂t}\right)d{x}_y+\left(\frac{∂f}{∂s}-\frac{∂f}{∂t}\right)\left(0\right) \\ dw=\left(\frac{∂f}{∂x}+2\frac{∂f}{∂s}+2\frac{∂f}{∂t}\right)d{x}_y \end{gathered}$$

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

Next divide by $d{x}_y$, then:

${\left(\frac{∂w}{∂x}\right)}_y={\left(\frac{∂f}{∂x}\right)}_{s,t}+2{\left(\frac{∂f}{∂s}\right)}_{t,x}+2{\left(\frac{∂f}{∂t}\right)}_{x,s}$

Substitute the values of$f_1$,$f_2$ , and $f_3$as:

$$\begin{gathered} {\left(\frac{∂w}{∂x}\right)}_y={\left(\frac{∂f}{∂x}\right)}_{s,t}+2{\left(\frac{∂f}{∂s}\right)}_{t,x}+2{\left(\frac{∂f}{∂t}\right)}_{x,s} \\ {\left(\frac{∂w}{∂x}\right)}_y=f_1+2f_2+2f_3 \end{gathered}$$

Therefore, the answer is ${\left(\frac{∂w}{∂x}\right)}_y=f_1+2f_2+2f_3$.