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Explain why Theorem 12.33 is a special case of Theorem

12.34 with n=2 and m=2

Short Answer

Expert verified

The required answer is

∂z∂t2=∂z∂x1∂x1∂t2+∂z∂x2∂x2∂t2

Step by step solution

01

Given information

The complete version of the chain rule is for a given function z=fx1,x2,…,xnand xi=uit1,t2,…,tmfor1≤i≤nor all values of t1,t2,…,tmat that uieach is differentiable and if f is differentiable at

(x1,x2,....xn)then

dzdtj=dzdx1dx1dtj+dzdx2dx2dtj+....+dzdxndxndtj......(1)

Where 1≤j≤m

The chain rule version three states that for a given function z=fx1,x2

and xi=uit1,t2for 1≤i≤2for all values of t1,t2at that each ui

is differentiable and if f is differentiable at x1,x2then

∂z∂t1=∂z∂x1∂x1∂t1+∂z∂x2∂x2∂t1And,∂z∂t2=∂z∂x1∂x1∂t2+∂z∂x2∂x2∂t2

02

The objective is to show when n=2 and m=2 then the complete version of the chain rule gives the chain rule version three.

When n=m=2then the complete version of the chain rule is as follows. For a given function z=fx1,x2and xi=uit1,t2for 1≤i≤2

For the values of t1andt2at which u1andu2are differentiable and if

fis differentiable at x1andx2then

The first put n=2and m=1in the equation (1)

∂z∂t1=∂z∂x1∂x1∂tt+∂z∂x2∂x2∂t1

Again put n=2and m=2in the equation (1)

∂z∂t2=∂z∂x1∂x1∂t2+∂z∂x2∂x2∂t2

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