Skills Review for Stokes’ Theorem and the Divergence Theorem

Learning Outcomes

  • Apply basic derivative rules
  • Use the product rule for finding the derivative of a product of functions
  • Use the quotient rule for finding the derivative of a quotient of functions
  • Apply the chain rule together with the power and product rule
  • Evaluate indefinite integrals

In the section about Stokes’ Theorem, we will learn about Stokes’ Theorem, a higher-dimensional generalization of Green’s Theorem. Then, we will explore the Divergence Theorem. Here we will review various differentiation techniques and rules along with basic integration techniques.

Basic Derivative Rules

(See Module 6, Skills Review for Line Integrals and Conservative Vector Fields)

The Product Rule

(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)
Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function [latex]f(x)=x^2[/latex], whose derivative is [latex]f^{\prime}(x)=2x[/latex] and not [latex]\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1[/latex].

Product Rule

Let [latex]f(x)[/latex] and [latex]g(x)[/latex] be differentiable functions. Then

[latex]\frac{d}{dx}(f(x)g(x))=\frac{d}{dx}(f(x))\cdot g(x)+\frac{d}{dx}(g(x))\cdot f(x)[/latex]

 

That is,

if [latex]j(x)=f(x)g(x)[/latex] then [latex]j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)[/latex]

 

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Example: Applying the Product Rule to Binomials

For [latex]j(x)=(x^2+2)(3x^3-5x)[/latex], find [latex]j^{\prime}(x)[/latex] by applying the product rule.

Try It

Use the product rule to obtain the derivative of [latex]j(x)=2x^5(4x^2+x)[/latex].

The Quotient Rule

(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

[latex]\frac{d}{dx}(x^2)=2x[/latex], which is not the same as [latex]\dfrac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} =\dfrac{3x^2}{1}=3x^2[/latex]

 

The Quotient Rule

Let [latex]f(x)[/latex] and [latex]g(x)[/latex] be differentiable functions. Then

[latex]\frac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\frac{d}{dx}(f(x))\cdot g(x)-\dfrac{d}{dx}(g(x))\cdot f(x)}{(g(x))^2}[/latex]

 

That is,

if [latex]j(x)=\dfrac{f(x)}{g(x)}[/latex], then [latex]j^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}[/latex]

 

Example: Applying the Quotient Rule

Use the quotient rule to find the derivative of [latex]k(x)=\dfrac{5x^2}{4x+3}[/latex]

Try It

Find the derivative of [latex]h(x)=\dfrac{3x+1}{4x-3}[/latex]

The Chain Rule

(See Module 6, Skills Review for Line Integrals and Conservative Vector Fields)

Indefinite Integrals

(also in Module 3, Skills Review for Calculus of Vector-Valued Functions)

Power Rule for Integrals

For [latex]n \ne −1[/latex],

[latex]\displaystyle\int x^n dx=\dfrac{x^{n+1}}{n+1}+C[/latex]

 

Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in Appendix B: Table of Derivatives.

Integration Formulas
Differentiation Formula Indefinite Integral
[latex]\frac{d}{dx}(k)=0[/latex] [latex]\displaystyle\int kdx=\displaystyle\int kx^0 dx=kx+C[/latex]
[latex]\frac{d}{dx}(x^n)=nx^{n-1}[/latex] [latex]\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C[/latex] for [latex]n\ne −1[/latex]
[latex]\frac{d}{dx}(\ln |x|)=\frac{1}{x}[/latex] [latex]\displaystyle\int \frac{1}{x}dx=\ln |x|+C[/latex]
[latex]\frac{d}{dx}(e^x)=e^x[/latex] [latex]\displaystyle\int e^x dx=e^x+C[/latex]
[latex]\frac{d}{dx}(\sin x)= \cos x[/latex] [latex]\displaystyle\int \cos x dx= \sin x+C[/latex]
[latex]\frac{d}{dx}(\cos x)=− \sin x[/latex] [latex]\displaystyle\int \sin x dx=− \cos x+C[/latex]
[latex]\frac{d}{dx}(\tan x)= \sec^2 x[/latex] [latex]\displaystyle\int \sec^2 x dx= \tan x+C[/latex]
[latex]\frac{d}{dx}(\csc x)=−\csc x \cot x[/latex] [latex]\displaystyle\int \csc x \cot x dx=−\csc x+C[/latex]
[latex]\frac{d}{dx}(\sec x)= \sec x \tan x[/latex] [latex]\displaystyle\int \sec x \tan x dx= \sec x+C[/latex]
[latex]\frac{d}{dx}(\cot x)=−\csc^2 x[/latex] [latex]\displaystyle\int \csc^2 x dx=−\cot x+C[/latex]
[latex]\frac{d}{dx}( \sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}[/latex] [latex]\displaystyle\int \frac{1}{\sqrt{1-x^2}} dx= \sin^{-1} x+C[/latex]
[latex]\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}[/latex] [latex]\displaystyle\int \frac{1}{1+x^2} dx= \tan^{-1} x+C[/latex]
[latex]\frac{d}{dx}(\sec^{-1} |x|)=\frac{1}{x\sqrt{x^2-1}}[/latex] [latex]\displaystyle\int \frac{1}{x\sqrt{x^2-1}} dx= \sec^{-1} |x|+C[/latex]

Example: Evaluating Indefinite Integrals

Evaluate each of the following indefinite integrals:

  1. [latex]\displaystyle\int (5x^3-7x^2+3x+4) dx[/latex]
  2. [latex]\displaystyle\int \frac{x^2+4\sqrt[3]{x}}{x} dx[/latex]
  3. [latex]\displaystyle\int \frac{4}{1+x^2} dx[/latex]
  4. [latex]\displaystyle\int \tan x \cos x dx[/latex]

Try It

Evaluate [latex]\displaystyle\int (4x^3-5x^2+x-7) dx[/latex]

Try It