Q. 5. - Calculus

Question

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 required answer is 

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

(Since the function is of a single variable so partial derivative

is the same as a normal derivative)

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

$t_1,t_2,\dots ,t_m$at which each $u_i$is differentiable and if $f$  is differentiable at $\left(x_1,x_2,\dots ,x_{n }\right)$then

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

Where $1\le j\le m$

Step 2: The objective is to show when n = 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 $z=f\left(x_1\right)$and $x_i=u_i\left(t_1\right)$for $1\le i$for the values of

$t_1$ at which $u_1$is differentiable and if $f$  is differentiable at $x_1$

Then put $n=m=1$in the equation $\left(1\right)$

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