91Ó°ÊÓ

It is given that

$z=f(x,y),x=u(s,t)\text{and}y=v(s,t)$

$u,v$ are differentiable for all values of$s,t$

Let $z=f(x,y)$be two variable function, $f(x,y)$is said to be differentiable at $\left(x_0,y_0\right)$if partial derivative of $f_x\left(x_0,y_0\right)\text{and}{f}_y\left(x_0,y_0\right)$both exists and

$\mathrm{Δ}z=f_x\left(x_0,y_0\right)\mathrm{Δ}x+f_y\left(x_0,y_0\right)\mathrm{Δ}y+{\mathrm{Î\mu }}_1\mathrm{Δ}x+{\mathrm{Î\mu }}_2\mathrm{Δ}y$

${\mathrm{Î\mu }}_1\text{and}{\mathrm{Î\mu }}_2$are functions of $\mathrm{Δ}x\text{and}\mathrm{Δ}y$ as$(\mathrm{Δ}x,\mathrm{Δ}y)→(0,0)$

Since $z$is differentiable

$\mathrm{Δ}z=z_x\mathrm{Δ}x+z_y\mathrm{Δ}y+{\mathrm{Î\mu }}_1\mathrm{Δ}x+{\mathrm{Î\mu }}_2\mathrm{Δ}y$

$=\left(z_x+{\mathrm{Î\mu }}_1\right)\mathrm{Δ}x+\left(z_y+{\mathrm{Î\mu }}_2\right)\mathrm{Δ}y$

Also $x,y$are differentiable for all $s,t$

$\mathrm{Δ}x=x_s\mathrm{Δ}s+x_1\mathrm{Δ}t+{\mathrm{Î\mu }}_3\mathrm{Δ}s+{\mathrm{Î\mu }}_4\mathrm{Δ}t$

${\mathrm{Î\mu }}_3→0,{\mathrm{Î\mu }}_4→0\text{as}(\mathrm{Δ}s,\mathrm{Δ}t)→(0,0)$

Also

$\mathrm{Δ}y=y_s\mathrm{Δ}s+y_t\mathrm{Δ}t+{\mathrm{Î\mu }}_5\mathrm{Δ}s+{\mathrm{Î\mu }}_6\mathrm{Δ}t$

${\mathrm{Î\mu }}_5→0,{\mathrm{Î\mu }}_6→0\text{as}(\mathrm{Δ}s,\mathrm{Δ}t)→(0,0)$

Hence, $\mathrm{Δ}z=\left(z_x+{\mathrm{Î\mu }}_1\right)\mathrm{Δ}x+\left(z_y+{\mathrm{Î\mu }}_2\right)\mathrm{Δ}y$

$=\left(z_x+{\mathrm{Î\mu }}_1\right)\left(x_s\mathrm{Δ}s+x_t\mathrm{Δ}t+{\mathrm{Î\mu }}_3\mathrm{Δ}s+{\mathrm{Î\mu }}_4\mathrm{Δ}t\right)+\left(z_y+{\mathrm{Î\mu }}_2\right)\left(y_s\mathrm{Δ}s+y_t\mathrm{Δ}t+{\mathrm{Î\mu }}_5\mathrm{Δ}s+{\mathrm{Î\mu }}_6\mathrm{Δ}t\right)$

Solving, we get

$=\left(z_x{x}_s+z_y{y}_s\right)\mathrm{Δ}s+\left(z_x{x}_t+z_y{y}_t\right)\mathrm{Δ}t+{\mathrm{Î\mu }}_7\mathrm{Δ}s+{\mathrm{Î\mu }}_8\mathrm{Δ}t$

where ${\mathrm{Î\mu }}_7=z_x{\mathrm{Î\mu }}_3+z_y{\mathrm{Î\mu }}_5+{\mathrm{Î\mu }}_1{x}_s+{\mathrm{Î\mu }}_2{y}_s+{\mathrm{Î\mu }}_1{\mathrm{Î\mu }}_3+{\mathrm{Î\mu }}_2{\mathrm{Î\mu }}_5$

$=\left(z_x+{\mathrm{Î\mu }}_1\right){\mathrm{Î\mu }}_3+{\mathrm{Î\mu }}_1{x}_s+{\mathrm{Î\mu }}_2{y}_s+\left(z_y+{\mathrm{Î\mu }}_2\right){\mathrm{Î\mu }}_5$

and

${\mathrm{Î\mu }}_8=z_x{\mathrm{Î\mu }}_4+z_y{\mathrm{Î\mu }}_6+{\mathrm{Î\mu }}_1{x}_t+{\mathrm{Î\mu }}_2{y}_t+{\mathrm{Î\mu }}_1{\mathrm{Î\mu }}_4+{\mathrm{Î\mu }}_2{\mathrm{Î\mu }}_6$

$=\left(z_x+{\mathrm{Î\mu }}_1\right){\mathrm{Î\mu }}_4+{\mathrm{Î\mu }}_1{x}_t+{\mathrm{Î\mu }}_2{y}_t+\left(z_y+{\mathrm{Î\mu }}_2\right){\mathrm{Î\mu }}_6$

If $(\mathrm{Δ}s,\mathrm{Δ}t)→(0,0)$, ${\mathrm{Î\mu }}_7→0\text{and}{\mathrm{Î\mu }}_8→0$

Solving further,

$\frac{\mathrm{Δ}z}{\mathrm{Δ}s}=\frac{1}{\mathrm{Δ}s}\left\{\left(z_x{x}_s+z_y{y}_s\right)\mathrm{Δ}s+\left(z_x{x}_t+z_y{y}_t\right)·0+{\mathrm{Î\mu }}_7\mathrm{Δ}s+{\mathrm{Î\mu }}_8·0\right\}$

$=z_x{x}_s+z_y{y}_s+{\mathrm{Î\mu }}_7$

Taking limit on both sides

$\underset{\mathrm{Δ}x→0}{\lim }\frac{\mathrm{Δ}z}{\mathrm{Δ}s}=\underset{\mathrm{Δ}x→0}{\lim }\left(z_x{x}_s+z_y{y}_s+{\mathrm{Î\mu }}_7\right)$

$=z_x{x}_s+z_y{y}_s$

$\frac{∂z}{∂s}=\frac{∂z}{∂x}\frac{∂x}{∂s}+\frac{∂z}{∂y}\frac{∂y}{∂s}$

Taking partial differentiation of$z$wrt$y$,

$=z_x{x}_t+z_y{y}_t+{\mathrm{Î\mu }}_8$

Taking limit on both sides

$\underset{\mathrm{Δ}y→0}{\lim }\frac{\mathrm{Δ}z}{\mathrm{Δ}t}=\underset{\mathrm{Δ}y→0}{\lim }\left(z_x{x}_t+z_y{y}_t+{\mathrm{Î\mu }}_8\right)$

$=z_x{x}_t+z_y{y}_t$

$\frac{∂z}{∂t}=\frac{∂z}{∂x}\frac{∂x}{∂t}+\frac{∂z}{∂y}\frac{∂y}{∂t}$

Hence proved.