Q. 6 - Calculus

Question

6. Explain why Theorem $12.32$ is a special case of Theorem $12.34$ with $n=2$ and $m=1$.

Step-by-Step Solution

Answer

The theorem 12.32 is determined to be a special case with $n=2 and m=1$

Introduction

The given is the points $n=2 and m=1$

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

Step 1

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 $1\le I\le n$

$\left(x_1,x_2,\dots ,x_n\right)$ 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 $1\le j\le m$.

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

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

Step 2

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 $1\le i\le 2$, 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)

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

Hence proved.