Problem 37 - Calculus Early Transcendentals

Question

Show that the operation of taking the gradient of a function has the given property. Assume that \(u\) and \(v\) are differentiable functions of \(x\) and \(y\) and that \(a, b\) are constants. $$\begin{array}{ll}{\text { (a) } \nabla(a u+b v)=a \nabla u+b \nabla v} & {\text { (b) } \nabla(u v)=u \nabla v+v \nabla u} \\ {\text { (c) } \nabla\left(\frac{u}{v}\right)=\frac{v \nabla u-u \nabla v}{v^{2}}} & {\text { (d) } \nabla u^{n}=n u^{a-1} \nabla u}\end{array}$$

Step-by-Step Solution

Answer

All gradient properties (a) through (d) are valid based on calculus rules.

Step 1: Understand the Problem

The problem involves verifying that the gradient operator \( abla \) satisfies certain algebraic properties when applied to differentiable functions \( u \) and \( v \). We need to show the validity of the gradient operation properties (a) through (d) using calculus.

Step 2: Prove Property (a)

The gradient of a function \( f(x, y) \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). To prove \( abla(a u+b v)=a abla u+b abla v \), compute:1. \( abla u = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} \right) \) and \( abla v = \left( \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y} \right) \).2. \( abla(au + bv) = \left( a\frac{\partial u}{\partial x} + b\frac{\partial v}{\partial x}, a\frac{\partial u}{\partial y} + b\frac{\partial v}{\partial y} \right) \).3. Since this equals \( a abla u + b abla v \), property (a) holds.

Step 3: Prove Property (b)

For \( abla(u v) = u abla v + v abla u \), apply the product rule for differentiation:1. \( abla(uv) = \left( \frac{\partial (uv)}{\partial x}, \frac{\partial (uv)}{\partial y} \right) \).2. By the product rule: \( \frac{\partial (uv)}{\partial x} = u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x} \) and \( \frac{\partial (uv)}{\partial y} = u \frac{\partial v}{\partial y} + v \frac{\partial u}{\partial y} \).3. Therefore, \( abla (uv) = u abla v + v abla u \) holds true.

Step 4: Prove Property (c)

For \( abla\left(\frac{u}{v}\right)=\frac{v abla u-u abla v}{v^{2}} \), use the quotient rule:1. \( abla\left(\frac{u}{v}\right) = \left( \frac{\partial (u/v)}{\partial x}, \frac{\partial (u/v)}{\partial y} \right) \).2. Applying the quotient rule: \( \frac{\partial}{\partial x}\left( \frac{u}{v} \right) = \frac{v\frac{\partial u}{\partial x} - u\frac{\partial v}{\partial x}}{v^2} \) and similarly for \( \frac{\partial}{\partial y} \).3. Thus, \( abla\left(\frac{u}{v}\right)=\frac{v abla u-u abla v}{v^{2}} \) is valid.

Step 5: Prove Property (d)

For \( abla u^{n}=n u^{n-1} abla u \), use the chain rule of differentiation:1. \( abla u^n = \left( \frac{\partial u^n}{\partial x}, \frac{\partial u^n}{\partial y} \right) \).2. By chain rule: \( \frac{\partial u^n}{\partial x} = n u^{n-1} \frac{\partial u}{\partial x} \) and \( \frac{\partial u^n}{\partial y} = n u^{n-1} \frac{\partial u}{\partial y} \).3. So, \( abla u^{n}=n u^{n-1} abla u \) is indeed correct.

Key Concepts

Differentiable Functions

Differentiable functions are foundational to calculus and are often prerequisites for defining gradients, derivatives, and various rules of differentiation. A function is said to be differentiable at a point when its derivative exists at that point. In simpler terms, the graph of the function looks smooth at that point, without any breaks or sharp corners.
When dealing with functions of two variables, such as those found in this exercise, a function is considered differentiable if it possesses partial derivatives with respect to each of the variables that are continuous around the point of interest. This continuity and smoothness are crucial as they allow the use of gradient operators, making it possible to calculate how the function changes in space.
Remember, differentiability implies continuity, but not vice versa. Therefore, if a function is differentiable everywhere in its domain, it automatically suggests that it behaves well, enabling us to apply advanced calculus tools like gradients and the rules discussed later.

Product Rule

The product rule is a fundamental formula used for differentiating the product of two functions. If you have two differentiable functions, say \( u(x, y) \) and \( v(x, y) \), and you want to find the derivative of their product, the product rule is the perfect tool for this purpose.
The rule states that the gradient of the product of two functions is given by:

• \( abla(uv) = u abla v + v abla u \)

This means that to find the gradient of \( uv \), you take \( u \) times the gradient of \( v \), and add it to \( v \) times the gradient of \( u \).
This rule is extremely useful because it simplifies the process of differentiation when dealing with products of functions. It is crucial for solving complex problems involving the gradients of multiplied functions in higher dimensions.

Quotient Rule

When differentiating the quotient of two functions, the quotient rule comes into play. It is especially relevant when you're dealing with a function expressed as a ratio, such as \( \frac{u(x, y)}{v(x, y)} \). Both \( u \) and \( v \) must be differentiable, and \( v \) must be non-zero in the domain of interest.
The quotient rule for gradients is:

• \( abla\left(\frac{u}{v}\right) = \frac{v abla u - u abla v}{v^{2}} \)

This formula helps find the gradient of a quotient by calculating the differences in the gradients of the numerator and denominator, scaled by the square of the denominator. This approach efficiently manages the changes in compounded variables.
It's imperative to consistently apply the quotient rule correctly to navigate the intricacies involved in breaking down these differences, essential for accuracy in mathematical applications.

Chain Rule

The chain rule is vital when differentiating composite functions. For instance, if you have a function of the type \( u(x, y)^n \), the chain rule allows you to differentiate efficiently without directly expanding the function.

• The chain rule states: \( abla u^n = n u^{n-1} abla u \)

With this rule, you treat \( u \) as a single variable raised to a power , differentiate the outer function first, and then multiply by the derivative of the inner function, \( abla u \). This is particularly useful in multi-variable calculus where functions are embedded within each other.
Using the chain rule correctly helps avoid mistakes when dealing with nested functions, thereby maintaining the flow and ease of calculations.