Q. 6. - Calculus

Question

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 required answer is

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

Step 1: Given information

The complete version of the chain rule is for a given function $z=f\left(x_1,x_2,\dots ,x_n\right)$and $x_i=u_i\left(t_1,t_2,\dots ,t_m\right)$for $1\le i\le n$ for all values of $t_1,t_2,\dots ,t_m$at which each $u_i$ is differentiable and if $f$ is differentiable at

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

$\frac{\partial z}{\partial t_j}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_j}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_j}+\dots +\frac{\partial z}{\partial x_n}\frac{\partial x_n}{\partial t_j}\dots \dots \text{ (1) }$

Where $1\le j\le m$

The chain rule when $x$ is a single variable function is given by

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

Step 2: The objective is to show when n = 2   and m = 1   then the complete version of the chain rule gives the chain rule for a single variable.

When $n=m=1$then the complete version of the chain rule is as follows. For a given function $z=f\left(x_1,x_2\right)$and $x_i=u_i\left(t_1\right)$for $1\le i\le 2$

For the values of $t_1$ at which  are differentiable and if $f$

 is differentiable at $x_1 and x_2$ then $n=2$and $m=1$in the equation $\left(1\right)$

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