Question

If $\mathit{w}\mathbf{=}\mathit{f}\left(x,s,t\right)$, $\mathit{s}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathit{y}$, $\mathbf{}\mathit{t}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathit{y}$find ${\left(\partial w/\partial x\right)}_{\mathbf{y}}$in terms of$\mathit{f}$ and its derivatives.

Short answer

The value of ${\left(\frac{\partial w}{\partial x}\right)}_y$is $f_1+2f_2+2f_3$.

Step-by-step solution

Given Information

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 w=f(x,s,t)

Differentiate the function as:

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

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:

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

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

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

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

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

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

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

Suppose f1=(∂f∂x)s,t , f2=(∂f∂s)t,x,and  f3=(∂f∂t)x,s

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

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

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