91Ó°ÊÓ

The given is the points $n=2andm=2$

The objective is to prove that the theorem is a special case and explain why

The full version of the chain rule is for a specific function $z=f(x1,x2,...xn)andxi=ui(t1,t2...tm)$for $1≤i≤n$, if each $ui$is differentiable at all values of $t1,t2,...,tm$and if $f$is differentiable at all values of $t1,t2,...,tm$

$\left(x_1,x_2,…,x_n\right)$then

$\frac{∂z}{∂t_j}=\frac{∂z}{∂x_1}\frac{∂x_1}{∂t_j}+\frac{∂z}{∂x_2}\frac{∂x_2}{∂t_j}+…+\frac{∂z}{∂x_n}\frac{∂x_n}{∂t_j}……$

where $1≤j≤m$.

The third version of the chain rule 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 $1≤i≤2$, for all values of $t_1,t_2$ Each $ui$ is differentiable at $(x1,x2)$, and if $f$ is differentiable at $(x1,x2)$, then

And

The goal is to demonstrate that when $n=2$and $m=2$the complete version of the chain rule is chain rule version three.

When $n=m=2$,the full 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 $1≤i≤2$, for the values of $t_1$and $t_2$at which $u_1$and $u_2$are differentiable, and if $f$is differentiable at $x1$and $x2$, then put $n=2$and $m=1$in equation first.

Substitute $n=2$ and $m=2$ in equation (1)

Hence proved.