Explain why Theorem $12.33$ is a special case of Theorem

$12.34$ with $n=2$ and $m=2$

The complete version of the chain rule is for a given function $z=f\left(x_1,x_2,\dots ,x_n\right)$and $x_i=u_i\left(t_1,t_2,\dots ,t_m\right)$for$1\le i\le n$or all values of $t_1,t_2,\dots ,t_m$at that $u_i$each is differentiable and if $f$ is differentiable at

$(x_1,x_2,....x_n)$then

$\frac{dz}{d{t}_j}=\frac{dz}{d{x}_1}\frac{d{x}_1}{d{t}_j}+\frac{dz}{d{x}_2}\frac{d{x}_2}{d{t}_j}+....+\frac{dz}{d{x}_n}\frac{d{x}_n}{d{t}_j}......\left(1\right)$

Where $1\le j\le m$

The chain rule version three states that for a given function $z=f\left(x_1,x_2\right)$

and $x_i=u_i\left(t_1,t_2\right)$for $1\le i\le 2$for all values of $t_1,t_2$at that each $u_i$

is differentiable and if $f$ is differentiable at $\left(x_1,x_2\right)$then

$$\begin{gathered} \frac{\partial z}{\partial t_1}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_1}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_1} \\ And,\frac{\partial z}{\partial t_2}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_2}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_2} \end{gathered}$$

When $n=m=2$then the complete version of the chain rule is as follows. For a given function $z=f\left(x_1,x_2\right)$and $x_i=u_i\left(t_1,t_2\right)$for $1\le i\le 2$

For the values of $t_{1}and{t}_2$at which $u_{1}and{u}_2$are differentiable and if

$f$is differentiable at $x_{1}and{x}_2$then

The first put $n=2$and $m=1$in the equation $\left(1\right)$

$\frac{\partial z}{\partial t_1}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_t}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_1}$

Again put $n=2$and $m=2$in the equation $\left(1\right)$

$\frac{\partial z}{\partial t_2}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_2}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_2}$