Q. 7 - Calculus

Question

7. Explain why Theorem 12.33 is a special case of Theorem $12.34$ with $n=2$ and $m=2$.

Step-by-Step Solution

Answer

The theorem 12.33 is determined as a special case with the points $n=2 and m=2$

Introduction

The given is the points $n=2 and m=2$

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

Step 1

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

$\left(x_1,x_2,\dots ,x_n\right)$ then

$\frac{\partial z}{\partial t_j}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_j}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_j}+\dots +\frac{\partial z}{\partial x_n}\frac{\partial x_n}{\partial t_j}\dots \dots$

where $1\le j\le m$.

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 $1\le i\le 2$, for all values of $t_1,t_2$ Each $ui$ is differentiable at $(x1,x2)$, and if $f$ is differentiable at $(x1,x2)$, then

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

And

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

Step 2

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

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

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

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

Hence proved.