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 $f(x)=x^2$ , whose derivative is $f^{\prime}(x)=2x$ and not $\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1$ .

Product Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

\frac{d}{d x} (f (x) g (x)) = \frac{d}{d x} (f (x)) \cdot g (x) + \frac{d}{d x} (g (x)) \cdot f (x)

$\frac{d}{dx}(f(x)g(x))=\frac{d}{dx}(f(x))\cdot g(x)+\frac{d}{dx}(g(x))\cdot f(x)$

That is,

j (x) = f (x) g (x)

$j(x)=f(x)g(x)$ then

j^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x)

$j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)$

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 $j(x)=(x^2+2)(3x^3-5x)$ , find $j^{\prime}(x)$ by applying the product rule.

Show Solution

If we set $f(x)=x^2+2$ and $g(x)=3x^3-5x$ , then $f^{\prime}(x)=2x$ and $g^{\prime}(x)=9x^2-5$ . Thus,

j^{'} (x) = f^{'} (x) g (x) + g^{'} (x) f (x) = (2 x) (3 x^{3} - 5 x) + (9 x^{2} - 5) (x^{2} + 2)

$j^{\prime}(x)=f^{\prime}(x)g(x)+g^{\prime}(x)f(x)=(2x)(3x^3-5x)+(9x^2-5)(x^2+2)$ .

Simplifying, we have

j^{'} (x) = 15 x^{4} + 3 x^{2} - 10

$j^{\prime}(x)=15x^4+3x^2-10$ .

Try It

Use the product rule to obtain the derivative of $j(x)=2x^5(4x^2+x)$ .

Hint

Set $f(x)=2x^5$ and $g(x)=4x^2+x$ and use the preceding example as a guide.

Show Solution

$j^{\prime}(x)=10x^4(4x^2+x)+(8x+1)(2x^5)=56x^6+12x^5$ .

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

\frac{d}{d x} (x^{2}) = 2 x

$\frac{d}{dx}(x^2)=2x$ , which is not the same as

\frac{\frac{d}{d x} (x^{3})}{\frac{d}{d x} (x)} = \frac{3 x^{2}}{1} = 3 x^{2}

$\dfrac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} =\dfrac{3x^2}{1}=3x^2$

The Quotient Rule

Let $f(x)$ and $g(x)$ be differentiable functions. Then

\frac{d}{d x} (\frac{f (x)}{g (x)}) = \frac{\frac{d}{d x} (f (x)) \cdot g (x) - \frac{d}{d x} (g (x)) \cdot f (x)}{(g (x))^{2}}

$\frac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\frac{d}{dx}(f(x))\cdot g(x)-\dfrac{d}{dx}(g(x))\cdot f(x)}{(g(x))^2}$

That is,

j (x) = \frac{f (x)}{g (x)}

$j(x)=\dfrac{f(x)}{g(x)}$ , then

j^{'} (x) = \frac{f^{'} (x) g (x) - g^{'} (x) f (x)}{(g (x))^{2}}

$j^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}$

Example: Applying the Quotient Rule

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

Show Solution

Let $f(x)=5x^2$ and $g(x)=4x+3$ . Thus, $f^{\prime}(x)=10x$ and $g^{\prime}(x)=4$ . Substituting into the quotient rule, we have

k^{'} (x) = \frac{f^{'} (x) g (x) - g^{'} (x) f (x)}{(g (x))^{2}} = \frac{10 x (4 x + 3) - 4 (5 x^{2})}{(4 x + 3)^{2}}

$k^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}=\dfrac{10x(4x+3)-4(5x^2)}{(4x+3)^2}$ .

Simplifying, we obtain

k^{'} (x) = \frac{20 x^{2} + 30 x}{(4 x + 3)^{2}}

$k^{\prime}(x)=\dfrac{20x^2+30x}{(4x+3)^2}$ .

Try It

Find the derivative of $h(x)=\dfrac{3x+1}{4x-3}$

Hint

Apply the quotient rule with $f(x)=3x+1$ and $g(x)=4x-3$ .

Show Solution

$k^{\prime}(x)=-\dfrac{13}{(4x-3)^2}$ .

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 $n \ne −1$ ,

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

$\displaystyle\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$

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
$\frac{d}{dx}(k)=0$	$\displaystyle\int kdx=\displaystyle\int kx^0 dx=kx+C$
$\frac{d}{dx}(x^n)=nx^{n-1}$	$\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C$ for $n\ne −1$
$\frac{d}{dx}(\ln \|x\|)=\frac{1}{x}$	$\displaystyle\int \frac{1}{x}dx=\ln \|x\|+C$
$\frac{d}{dx}(e^x)=e^x$	$\displaystyle\int e^x dx=e^x+C$
$\frac{d}{dx}(\sin x)= \cos x$	$\displaystyle\int \cos x dx= \sin x+C$
$\frac{d}{dx}(\cos x)=− \sin x$	$\displaystyle\int \sin x dx=− \cos x+C$
$\frac{d}{dx}(\tan x)= \sec^2 x$	$\displaystyle\int \sec^2 x dx= \tan x+C$
$\frac{d}{dx}(\csc x)=−\csc x \cot x$	$\displaystyle\int \csc x \cot x dx=−\csc x+C$
$\frac{d}{dx}(\sec x)= \sec x \tan x$	$\displaystyle\int \sec x \tan x dx= \sec x+C$
$\frac{d}{dx}(\cot x)=−\csc^2 x$	$\displaystyle\int \csc^2 x dx=−\cot x+C$
$\frac{d}{dx}( \sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}$	$\displaystyle\int \frac{1}{\sqrt{1-x^2}} dx= \sin^{-1} x+C$
$\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^2}$	$\displaystyle\int \frac{1}{1+x^2} dx= \tan^{-1} x+C$
$\frac{d}{dx}(\sec^{-1} \|x\|)=\frac{1}{x\sqrt{x^2-1}}$	$\displaystyle\int \frac{1}{x\sqrt{x^2-1}} dx= \sec^{-1} \|x\|+C$

Example: Evaluating Indefinite Integrals

Evaluate each of the following indefinite integrals:

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

Show Solution

Using the properties of indefinite integrals, we can integrate each of the four terms in the integrand separately. We obtain
$\displaystyle\int (5x^3-7x^2+3x+4) dx=\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx$

From the Constant Multiples property of indefinite integrals, each coefficient can be written in front of the integral sign, which gives

$\displaystyle\int 5x^3 dx-\displaystyle\int 7x^2 dx+\displaystyle\int 3x dx+\displaystyle\int 4 dx=5\displaystyle\int x^3 dx-7\displaystyle\int x^2 dx+3\displaystyle\int x dx+4\displaystyle\int 1 dx$

Using the power rule for integrals, we conclude that

$\displaystyle\int (5x^3-7x^2+3x+4) dx=\frac{5}{4}x^4-\frac{7}{3}x^3+\frac{3}{2}x^2+4x+C$
Rewrite the integrand as
$\frac{x^2+4\sqrt[3]{x}}{x}=\frac{x^2}{x}+\frac{4\sqrt[3]{x}}{x}$

Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have

$\begin{array}{ll} \displaystyle\int (x+\frac{4}{x^{2/3}}) dx & =\displaystyle\int x dx+4\displaystyle\int x^{-2/3} dx \\ & =\frac{1}{2}x^2+4\frac{1}{(\frac{-2}{3})+1}x^{(-2/3)+1}+C \\ & =\frac{1}{2}x^2+12x^{1/3}+C \end{array}$
Using the properties of indefinite integrals, write the integral as
$4\displaystyle\int \frac{1}{1+x^2} dx$ .

Then, use the fact that $\tan^{-1} (x)$ is an antiderivative of $\frac{1}{1+x^2}$ to conclude that

$\displaystyle\int \frac{4}{1+x^2} dx=4 \tan^{-1} (x)+C$
Rewrite the integrand as
$\tan x \cos x=\frac{ \sin x}{ \cos x} \cos x= \sin x$ .

Therefore,

$\displaystyle\int \tan x \cos x dx=\displaystyle\int \sin x dx=− \cos x+C$

Try It

Evaluate $\displaystyle\int (4x^3-5x^2+x-7) dx$

Hint

Show Solution

$x^4-\frac{5}{3}x^3+\frac{1}{2}x^2-7x+C$

Vector Calculus: Prerequisite Material