91Ó°ÊÓ

If any function $v=v\left(x,y\right)$has independent variables which are also functions of some independent variables such that $x=x\left(s,t\right)\text{and}y=y\left(s,t\right)$, then chain rule for the function is given by:

$$\begin{gathered} \frac{\mathbf{∂}\mathbf{v}}{\mathbf{∂}\mathbf{s}}\mathbf{=}\frac{\mathbf{∂}\mathbf{v}}{\mathbf{∂}\mathbf{x}}\frac{\mathbf{∂}\mathbf{x}}{\mathbf{∂}\mathbf{s}}\mathbf{+}\frac{\mathbf{∂}\mathbf{v}}{\mathbf{∂}\mathbf{y}}\frac{\mathbf{∂}\mathbf{y}}{\mathbf{∂}\mathbf{s}} \\ \frac{\mathbf{∂}\mathbf{v}}{\mathbf{∂}\mathbf{t}}\mathbf{=}\frac{\mathbf{∂}\mathbf{v}}{\mathbf{∂}\mathbf{x}}\frac{\mathbf{∂}\mathbf{x}}{\mathbf{∂}\mathbf{t}}\mathbf{+}\frac{\mathbf{∂}\mathbf{v}}{\mathbf{∂}\mathbf{y}}\frac{\mathbf{∂}\mathbf{y}}{\mathbf{∂}\mathbf{t}} \end{gathered}$$

The given functions are: $x{s}^2+y{t}^2=1$and $x^{2}s+y^{2}t=xy-4$.

Differentiate$x{s}^2+y{t}^2=1$ as follows:

$$\begin{gathered} x{s}^2+y{t}^2=1 \\ 2sxds+s^{2}dx+2tydt+t^{2}dy=0 \\ {s}^{2}dx+t^{2}dy=-2tydt-2sxds \end{gathered}$$ ….. (1)

Similarly, differentiate$x^{2}s+y^{2}t=xy-4$ as follows:

$$\begin{gathered} x{s}^2+y{t}^2=xy-4 \\ 2xsdx+x^{2}ds+2ytdy+y^{2}dt=xdy+ydx \\ \left(y-2sx\right)dx-\left(x-2ty\right)dt=x^{2}dx+y^{2}dt \end{gathered}$$ ….. (2)

From equations (1) and (2):

$$\begin{gathered} dx=\left|\frac{\begin{array}{cc}-2tydt-2sxds& t^2\\ x^{2}ds+y^{2}dt& (x-2ty)\end{array}}{\begin{array}{cc}{s}^2& t^2\\ (y-2sx)& (x-2ty)\end{array}}\right| \\ =\frac{(-2s{x}^2+4xyst)ds+(-2xyt+4t^2{y}^2)dt-t^2{x}^{2}ds-t^2{y}^{2}dt}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \\ =\frac{(-2s{x}^2+4xyst)-t^2{x}^2}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}ds+\frac{(-2xyt+3t^2{y}^2)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}dt \end{gathered}$$

So, the partial derivatives are:

$$\begin{gathered} \frac{∂x}{∂s}=\frac{(-2s{x}^2+4xyst)-t^2{x}^2}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \\ \frac{∂x}{∂t}=\frac{(-2xy{t}^2+3t^2{y}^2)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \end{gathered}$$

At point $(x,y,s,t)=(1,-3,2,1)$, we get:

localid="1664255450661" $$\begin{gathered} {\left(\frac{∂x}{∂s}\right)}_{(x,y,s,t)}={\left[\frac{(-2s{x}^2+4xyst)-t^2{x}^2}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}\right]}_{(1,-3,2,1)} \\ =-\frac{19}{13} \end{gathered}$$

And,

$$\begin{gathered} {\left(\frac{∂x}{∂t}\right)}_{(x,y,s,t)}={\left[\frac{(-2xyt+3t^2{y}^2)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}\right]}_{(1,-3,2,1)} \\ =-\frac{21}{13} \end{gathered}$$

Again, from equations (1) and (2) solve as:

$$\begin{gathered} dx=\left|\frac{\begin{array}{cc}{s}^2& -2tydt-2sxds\\ y-2sx& -x^{2}ds-y^{2}dt\end{array}}{\begin{array}{cc}{s}^2& t^2\\ (y-2sx)& (x-2ty)\end{array}}\right| \\ =\frac{(x^2+2sxy-4x^2{s}^2)ds+(-2xyst-s^2{y}^2+y^{2}t)dt}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \\ =\frac{(x^2+2sxy-4x^2{s}^2)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}ds+\frac{(-2xyst-s^2{y}^2+y^{2}t)dt}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \end{gathered}$$

So, the partial derivatives are:

$$\begin{gathered} \frac{∂y}{∂s}=\frac{(-x^2+2sxy-4x^2{s}^2)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \\ \frac{∂y}{∂t}=\frac{(-2xyst-s^2{y}^2+y^{2}t)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2} \end{gathered}$$

At point $(x,y,s,t)=(1,-3,2,1)$and solve as:

localid="1664256338700" $$\begin{gathered} {\left(\frac{∂y}{∂s}\right)}_{(x,y,s,t)}={\left[\frac{(-x^2+2sxy-4x^2{s}^2)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}\right]}_{(1,-3,2,1)} \\ =\frac{24}{13} \end{gathered}$$

Also,

$$\begin{gathered} {\left(\frac{∂y}{∂t}\right)}_{(x,y,s,t)}={\left[\frac{(-2xyst-s^2{y}^2+y^{2}t)}{s^{2}x-2s^{2}ty-t^{2}y+2sx{t}^2}\right]}_{(1,-3,2,1)} \\ =\frac{6}{13} \end{gathered}$$

Hence, the required values are:

$$\begin{gathered} \left(\frac{∂x}{∂s}\right)=-\frac{19}{13} \\ \left(\frac{∂x}{∂t}\right)=-\frac{21}{13} \\ \left(\frac{∂y}{∂s}\right)=\frac{24}{13} \\ \left(\frac{∂y}{∂t}\right)=\frac{6}{13} \end{gathered}$$