Chapter 12: Q. 6. (page 963)

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

Short Answer

Expert verified

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 by step solution

Given information

The complete version of the chain rule is for a given function $z = f (x_{1}, x_{2}, \dots, x_{n})$ and $x_{i} = u_{i} (t_{1}, t_{2}, \dots, t_{m})$ for $1 \leq i \leq n$ for all values of $t_{1}, t_{2}, \dots, t_{m}$ at which each $u_{i}$ is differentiable and if $f$ is differentiable at

$(x_{1}, x_{2}, \dots, x_{n})$

$\frac{\partial z}{\partial t_{j}} = \frac{\partial z}{\partial x_{1}} \frac{\partial x_{1}}{\partial t_{j}} + \frac{\partial z}{\partial x_{2}} \frac{\partial x_{2}}{\partial t_{j}} + \dots + \frac{\partial z}{\partial x_{n}} \frac{\partial x_{n}}{\partial t_{j}} \dots \dots (1)$

Where $1 \leq j \leq m$

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

$\frac{\partial z}{\partial t_{1}} = \frac{\partial z}{\partial x_{1}} \frac{\partial x_{1}}{\partial t_{1}} + \frac{\partial z}{\partial x_{2}} \frac{\partial x_{2}}{\partial t_{1}}$

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 $z = f (x_{1}, x_{2})$ and $x_{i} = u_{i} (t_{1})$ for $1 \leq i \leq 2$

For the values of $t_{1}$ at which are differentiable and if $f$

is differentiable at $x_{1} a n d x_{2}$ then $n = 2$ and $m = 1$ in the equation $(1)$

$\frac{\partial z}{\partial t_{1}} = \frac{\partial z}{\partial x_{1}} \frac{\partial x_{1}}{\partial t_{t}} + \frac{\partial z}{\partial x_{2}} \frac{\partial x_{2}}{\partial t_{1}}$

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

Short Answer

Step by step solution

Given information

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.

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Explain why Theorem12.32 is a special case of Theorem12.34 with n=2 andm=1

Short Answer

Step by step solution

Given information

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.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Pure Maths

Probability and Statistics

Calculus

Geometry

Applied Mathematics

Study anywhere. Anytime. Across all devices.

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