91影视

The given is the points $n=2andm=1$

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

For a given function, $z=f(x1,x2,...,xn)$is the complete version of chain rule. $xi=ui(t1,t2,...,tm)$and $xi=ui(t1,t2,...,tm)$for all values of $t1,t2,...,tm$at which each $ui$is differentiable and if $f$is differentiable at $1鈮�I鈮�n$

$\left(x_1,x_2,鈥�,x_n\right)$then

$\frac{鈭�z}{鈭�t_j}=\frac{鈭�z}{鈭�x_1}\frac{鈭�x_1}{鈭�t_j}+\frac{鈭�z}{鈭�x_2}\frac{鈭�x_2}{鈭�t_j}+鈥�+\frac{鈭�z}{鈭�x_n}\frac{鈭�x_n}{鈭�t_j}鈥�鈥�$

where $1鈮�j鈮�m$.

When $x$ is a single variable function, the chain rule is

The goal is to demonstrate that for $n=2$and $m=1$, the complete version of chain rule offers a single variable chain rule.

When $n=m=1$then the following is the whole version of the chain rule:. For the function $z=f\left(x_1,x_2\right)$ and $x_i=u_i\left(t_1\right)$ for $1鈮�i鈮�2$, for the values of $t_1$ at which $u_1$ is differentiable and if $f$ is differentiable at $x_1$ and $x_{2}$ then substitute $n=2$ and $m=1$ in equation ( 1)

Hence proved.