Q. 5 - Calculus

Question

5. Explain why the chain rule from Chapter 2 is a special case of Theorem $12.34$ with $n=1$ and $m=1$.

Step-by-Step Solution

Answer

The chain rule from theorem 12.34 with $n=1 and m=1$is determined to be proved

Introduction

The objective is to prove that the chain rule in chapter 2 is special case and to explain why

Step 1

For a given function, $z=f(x_1,x_2....,x_n)$ is the complete version of chain rule. $x_i=u_i(t_1,t_2,...t_m)$ and $t_1,t_2,...t_n$ for all values at which each $u_i$ 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 $1\le j\le m$.

For a single variable function, the chain rule is

$\frac{dz}{d{t}_1}=\frac{dz}{d{x}_1}\frac{d{x}_1}{d{t}_1}$

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

The complete version of the chain rule is as follows when $n=m=1$. $z=f\left(x_1\right)$ and $x_i=u_i\left(t_1\right)$ for a given function $n=m=1$ in place of $1\le i$ for the values of $t_1$ at which $u_1$ is differentiable and if $f$ is differentiable at $x_1$.

$\frac{dz}{d{t}_1}=\frac{dz}{d{x}_1}\frac{d{x}_1}{d{t}_1}$ [Because the function has only one variable, the partial derivative is identical to the normal derivative.]

Hence proved.