91Ó°ÊÓ

The given equations $w=f\left(x,x^2+y^2,2xy\right)$.

Rewrite the function$w=f\left(x,x^2+y^2,2xy\right)$ as follows:

$w=f\left(x,s,t\right)$

Where$s=x^2+y^2...\left(1\right)$ and $t=2xy...\left(2\right)$.

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:

$$\begin{gathered} ds=2xdx+2ydy \\ dt=2ydx+2xdy \end{gathered}$$

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

$$\begin{gathered} dw=\frac{∂f}{∂x}dx+\frac{∂f}{∂s}ds+\frac{∂f}{∂t}dt \\ dw=\frac{∂f}{∂x}dx+\frac{∂f}{∂s}\left(2xdx+2ydy\right)+\frac{∂f}{∂t}\left(2ydx+2xdy\right) \\ dw=\frac{∂f}{∂x}dx+2x\frac{∂f}{∂s}dx+2y\frac{∂f}{∂s}dy+2y\frac{∂f}{∂t}dx+2x\frac{∂f}{∂t}dy \\ dw=\left(\frac{∂f}{∂x}+2x\frac{∂f}{∂s}+2y\frac{∂f}{∂t}\right)dx+\left(2y\frac{∂f}{∂s}+2x\frac{∂f}{∂t}\right)dy...\left(4\right) \end{gathered}$$

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

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

role="math" localid="1664263726727" $$\begin{gathered} dw=\left(\frac{∂f}{∂x}+2x\frac{∂f}{∂s}+2y\frac{∂f}{∂t}\right)dx+\left(2y\frac{∂f}{∂s}+2x\frac{∂f}{∂t}\right)dy \\ d{w}_y=\left(\frac{∂f}{∂x}+2x\frac{∂f}{∂s}+2y\frac{∂f}{∂t}\right)d{x}_y+\left(2y\frac{∂f}{∂s}+2x\frac{∂f}{∂t}\right)\left(0\right) \\ d{w}_y=\left(\frac{∂f}{∂x}+2x\frac{∂f}{∂s}+2y\frac{∂f}{∂t}\right)d{x}_y \end{gathered}$$

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

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

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

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

$$\begin{gathered} {\left(\frac{∂w}{∂x}\right)}_y={\left(\frac{∂f}{∂x}\right)}_{s,t}+2x{\left(\frac{∂f}{∂s}\right)}_{t,x}+2y{\left(\frac{∂f}{∂t}\right)}_{x,s} \\ {\left(\frac{∂w}{∂x}\right)}_y=f_1+2x{f}_2+2y{f}_3 \end{gathered}$$

Therefore, the answer is${\left(\frac{∂w}{∂x}\right)}_y=f_1+2x{f}_2+2y{f}_3$ .