Calculus SG Impo

8/9/2019 Calculus SG Impo

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2008 Rose-Hulman Institute of Technologys Homework HotlinePage 1 o

Table of Contents

Derivatives 2Limits ........ 23Riemann Sums....28

Integrals..32

Page 1 of 37. Form 1 of 10, Front. File:Calculus-SG.pdf


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2008 Rose-Hul

The derivative of a function is th

particular point. For example, th

the point The slopeofpoint .

Notationsfor the derivative incl

We can define the derivative in t

point, and draw the line betweapproximately the slope of the fu

Now move those points closer to

closer to the slope of the functio

points move c loser and c loser to

an Institute of Technologys Homework Hotli

e instantaneous rate of change, or the slope of t

green line in the picture below is tangentto th

the green line in this picture is the derivative of

ude, ,

erms of limits. Suppose we take two points on

n them, called a secant line. The slope of the s

nction at the point .

, and draw another secant line. The slope of tat . If we take the limit of the slope of the seether, we get the definition of the derivative at

Slope=

e

Page 2 of 37

at function at a

e curve at at the

either side of a

cant line is

his line is much

cant line as these

.

Page 2 of 37. Form 1 of 10, Back. File:Calculus-SG.pdf


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lim +

The limit definition is the most basic formula for calculating the derivative of a function. In formula, is the function to be differentiated, and represents a small change in , whigoes to zero.Example:

Find the derivative of = . Then calculate the slope of at = 4 . =lim = l i m +

Change to .

= l i m

+ 2 +

FOIL the + . = l i m 2 +

The x terms cancel. = l i m 2 + The cancel. = goes to zero.To find the slope of at the point = 4, we plug in 4 to the derivative function. =24 = 24 = 8The slope of the line that is tangent to

when

= 4is

8.

Sometimes, it is not possible to find the derivative of a function at a certain point because thatfunction is not continuous.



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2008 Rose-Hulman Institute of Technologys Homework HotlinePage 4 of 37

Example: Find the derivative of at = 0.We can do this one using the limit definition (try it for practice!), or we can look at itgraphically:

The absolute value function has a derivative of 1, depending on the sign of . = 1 >0

1 < 0

But what is the derivative at = 0?When you studied limits, you probably learned that when the right-hand and left-hand limits at apoint dont agree, the limit at that point does not exist.

The derivative of does not exist at = 0because the function is not continuous there.

In many cases, the limit definition of the derivative can be bypassed. The followingdifferentiation rules show some of the shortcuts that can be taken.

The Power Rule

The power rule is one of the methods of taking the derivative that you will use most often, so you

should make yourself familiar with it.

For any function , the derivative can be found as:

Slope = 1Slope = -1

=



3cos 9sin3cos 19 sin = 19 csc .

19 cot+ .

19 9 + + cot Integration Shortcuts:

Just like rules of differentiation, there are rules you can follow to quickly integrate most

functions. Here is a list of the rules you should familiarize yourself with. Let , , and beconstants.

0 = = +

=

[ ] = = + n1 = sin + sin = cos + sec = tan + sectan = sec + csc = cot + csc cot =csc + Hint: Check your understanding by taking the derivative of the right side. You should get thintegrand on the left.



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Trigonometric Substitution:

Recall the Pythagorean Theorem and the identities of right-angle trigonometry.

Trigonometric substitution allows us to simplify radicals in the integrand of the forms , + , and to trigonometric functions.Using the reference triangles above, radicals in the integrand may be replaced as follows:

=

+ = s e c = t a n (positive if u > aand negative if u < -a)Example: Solution: Reference triangle

9 = 3 . = 3 sin .

= 3 c os ,

= 9sin

.9 = 3cos 9 .

+

9



= To apply this rule, multiply the coefficient, , by thecurrent exponent, ,and then decrease the exponent onby 1.

Examples: = 2 Multiply by2 , then decrease the exponent to 1. 2 = 10 Multiply by 5, then decrease the exponent to 4. 2 = 4

Multiply by -

2, then decrease the exponent to -

3.

The Constant Rule

The derivative of any constant is zero.

This can be observed graphically, because a line representing a constant function is horizonta

with zero slope. This can also be shown in terms of the power rule, because a constant canwritten as .Example:

5 = 5 = 0Derivative of a Sum

The derivative of a sum of terms is equal to the sum of the derivatives of the terms.

+ = + .This makes it easy to differentiate polynomials using the power rule.

Example: 2 + = 2 + = 6 + 7

Derivatives of Trigonometric Functions

The derivatives of common trigonometric functions should be memorized.

sinx=cos cos = sin x



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tan = sec =sec tan

cot =

csc = c s c cot

The Constant Multiple Rule

If a function is multiplied by a constant coefficient, you may take the derivative of the functionalone, and then multiply by the constant.

= Example:

3 sin = 3

sin = 3 cos

Differentiation Rules of Logarithmic and Exponential Functions

These commonly used derivatives should be memorized. Let be a constant. = = ln

ln = 1

= 1 ln



Tips for choosing and : The function which is easiest to integrate, or which simplifies the most when integrate

should be chosen as . The function chosen as should simplify when differentiated, and should not be toodifficult to differentiate. If the integration problem becomes more complicated when you apply integration by

parts, try picking and another way.Example:

ln Solution:

lnis hard to integrate,but easy to differentiate,so = ln , =

= = = 3 . = l n = 1 . l n 3

3 1

l n 3 13 . + .



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Example:

+ Solution:

= + 1 = 2 . + 1 = .12 2 12 2 . 12 = 2. + , . .6 + . + 16 + = + 1. .

Integration by Parts:

When integrating the product of two functions, , you may use integration by parts.This is a reverse version of the product rule. Let = , and = .

=

Often, it does matter which function you choose as and which is . The goal of integrationby parts is to reduce the integral to another integral, , which is simpler to evaluate.We dont want to make things more complicated.



The Product Rule

For two functions multiplied by one another, the product rule can be used:

= To apply this rule, differentiate each function separatelMultiply the first function by the derivative of the seco + and vice-versa. Then add the two terms together.

Example: Find the derivative of = sin. = , = sin . = 2, = . = + = cos + 2.

The Quotient Rule

For two terms divided by one another, the quotient rule is a convenient shortcut.

= Again, differentiate each function separately. Thenmultiply the numerator function by the derivative of thedenominator and vice versa. Subtract the two terms, asshown. Divide by the square of the denominator.

Example: Find the derivative of =

= cos , = 2. = sin , = 2.

= + = 2 2sin 4 = sin 2 .



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The Chain Rule

For a composition of two functions,

and , use the chain rule.

= To use this formula, first find the derivative of the insideand outside functions separately. Apply the derivative ofthe outside function to the inside function. Then multiplyby the derivative of the inside function.

Example: Find the derivative of 2 x + 1Inside function: 2 + 1 2 + 1 = 2Outside function: = 3

2 + 1

= 32 + 1

2 = 62 + 1

Differentiation Rules of Inverse Trigonometric Functions

It can be useful to memorize these less common derivatives, especially in preparation for findingintegrals in the next semester of calculus. You should learn to recognize them.

= 1 1 = 1 1

=1

1 +

= 11 +



Example: Find the area under the curve = 3 from = 1to = 2 . 3 .

3 = 3 = 2 1 = 7 .

When taking the integral of a function, you must apply the rules of differentiation in reverse.This can be tricky at first.

Example: Apply the reverse power rule to find the following indefinite integral:

2 First, think of a derivative that has an in it. I know that = 4.We need to turn that 4 into a 2, so introduce a factor of .Deduce that

= 2.Write the anti-derivative, and add .

2 = 12 + U-substitution:

You should recall the chain rule from differentiation. When the chain rule is applied to a

function , the derivative has the form .When integrating a function of the form

, we do the chain rule in reverse.

First, make the substitution = . Find = . Then = Make sure yointegral has the form . Then take the anti-derivative.



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Integration is the opposite of differentiation . For this reason, the terms integral and anti-derivative are sometimes interchanged. Integrals, like derivatives, can be calculated withouttaking a limit.

An integral of the form

, with upper and lower bounds, is called a definite integral. It

represents the area under the curve from to , and it is equal to a number.An integral of the form , without upper and lower bounds, is called an indefiniteintegral. It represents all functions that have the derivative , and is equal to a family ofcurves.

= + Since the derivative of a constant is zero, taking the anti-derivative of produces a familyof curves, + , where can be any constant.Example: What is 2 ?We know the derivative of is 2. But the derivatives of + 1 , + 7,and +69 are alsoall equal to

2. In fact, the derivative of

plus any number is

2.

2 = + .It is important to remember to add when taking an indefinite integral.

Definite integrals can be calculated from indefinite integrals using The Fundamental Theorem ofCalculus:

= Where

= + . (Notice that the

+ doesnt matter here, because when we

subtract + + , the s will cancel out.)



= 1 1 = 1 1Basics of Differentiation Practice Problems:

1.)Find, using the limit definition, the derivative of + 2.2.)Find the derivative of:

a.

b. 2c. 3d. 2 + 5e. 5 (Think of this as 5.)

3.)Use the product rule to find the derivative of:

a. (x+3)(x+4)

b.

+3x)4.)Use the quotient rule to find the derivative of:

a.

b.

5.)Use the chain rule to find the derivative of:

a. 7 + 1b. 3 + 6 + 2

Practice Problem Solutions:



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2008 Rose-Hul

Finding the Equation of th

Now that we know how to find t

line. Recall that to write the equ

get the slope from taking the deri

the function.

Example: Find the equation of th

= 2.

First, find the derivative of the f

=

+ 3

=

+

3

= 3 + 3

To find the slope at 2, plug =2 = 3 2+ 32= 15

Plug = 2 into the function to g2= 2 + 32 6

Now we have slope = 15at tWe can use this information to t

= 8 = 15 2 = 1522

We can now plot and the ta


e Tangent Line

e slope of a function, we can find the equation

tion of a line we need the slope and a point on t

ivative of the function, and we can get a point b

e tangent line to the function= + 3

nction.

6

6

2into the derivative.

et the y-value of the point.

8

e point 2,8.e write the equation of the line in point-slope f

ngent line = 15 22 on the same set of axe

f(x)

e

Page 10 of 37

of the tangent

he line. We can

plugging it into

6at the point

rm.

.



= 1.516 + 1.519.75 + 1.519 + 1.513.75=24+29.625+28.5+20.625=.In this example, the right-hand sum is a lower approximation. The left-hand sum is an upperapproximation. We can conclude that the true area is between 84.75 and 102.75.

Finding the Exact Area through I ntegrals:

When estimating the area under a curve using Riemann sums, we saw that using more rectangled to more accurate approximations. If we let the size of the rectangles get smaller and smalland the number of rectangles get larger and larger, the sum of the rectangles approaches the t

area under the curve. Taking the limit of a Riemann approximation as (the number ofrectangles) goes to infinity gives the exact area under the curve, also called a definite integral

= = lim + This integral is shown as the limit of a right-hand sum. A left-hand sum or midpoint sum wou

have the same limit as n approaches infinity, and would give the same integral.

In this integral, and are theupper and lower boundaries. The

function is the integrand.The at the end tells us that weare taking the integral with

respect to the variable .



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Example:

Use four rectangles to find an upper and lower approximation of the area under = 2 0 between = 2 and = 4First find the width of each rectangle. To do this, divide the total width of the interval by thenumber of rectangles you want.

= 424 = 64 = 32Next we need the heights of the rectangles. To find the right-hand sumwe need to use theheights at the right endpoints of each rectangle.

The right endpoints of the four rectangles are , 1 , 2 , 4The function is = 2 0 .

Height of Rectangle 1 = .5 =20.5 =19.75Height of Rectangle 2 = 1 = 2 0 1 =19Height of Rectangle 3 = 2.5 =202.5 =13.75Height of Rectangle 4 = 4 = 2 0 4 = 4

Add up the areas of all the rectangles. = = 1.519.75 + 1.519 + 1.513.75 + 1 . 5 4=29.625+28.5+20.625+6 =.Note: Since all the rectangles are of equal width, you could sum the heights first, and then

multiply once by the width.

To find the left-hand sumyou will need to use the heights at the left endpoints of each rectangle

The left endpoints of the four rectangles are 2, , 1 , 2 The function is = 2 0

Height of Rectangle 1 =

2 =202

= 16

Height of Rectangle 2 = .5 =200.5 =19.75Height of Rectangle 3 = 1 = 2 0 1 = 19Height of Rectangle 4 = 2.5 =202.5 = 13.75

Sum the area of all the rectangles =


2008 Rose-Hul

Implicit Differentiation

Sometimes it is necessary to find

For example:

We cannot just take the derivativ

will use implicit differentiation.

1.

Take the derivative of bo

the derivative of an expre

2.

When you encounter y, t

. (The reason for this

3.

Once you have taken all

Lets find the derivative of the fu

3 = sin

1

+ 2 3 = cos

=


the derivative of a function which is not in the

3 = sin

e of this function because it is not solved for y.

Here are the rules for implicit di fferentiation:

th sides. Use all of the same differentiation rule

ssion involving only x.

ke the derivative using all the same rules, but al

lies in the chain rule. is a function of .)

f the derivatives, solve for

.

nction above.

On the left, we have used th

Solve for

.

(2,8)

e

Page 11 o

form =

Therefore, w

s as when ta

so multiply

e product rul



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Practice: Try using implicit differentiation on the following functions:

1. + = 252. sin + = cos

Solutions:

1. + = 252 + 2 = 0 Power Rule and Implicit Differentiation. = Solve for .

2. sin + = cos + 1 + = 2 Implicit Differentiationcos + + c o s + = 2 cos Distribute cos + . cos + 2 cos = c o s + Rearrange the terms.

+2cos = sin cos + Factor out . =

sin cos + +2cos Solve for .

So far, we have only taken the first derivative. There are also second, third, fourth, and evenhigher order derivatives. We already know that the first derivative gives us the slope of thefunction. The second derivative describes the rate of change of the slope. We call this concavity.

Notations for the second derivative include , , or .To find higher order derivatives, you take derivatives successively. For example, to find thesecond derivative you would first take the first derivative of the function, and then take thederivative again. Lets do an example.

= +cos



How to Calculate Riemann Sums:

Step 1: Find the width of the rectangles: = a = the starting pointb = the ending pointn = number of rectangles you want to approximate with.

Step 2: Find the heights of the rectangles: = Each rectangle spans a range of x-values. Which do we plug in to calculate the heigh

Left-hand Estimation: Use the x-value at the left end of each rectangle to calculate heRight-hand Estimation: Use the x-value at the right of each rectangle to calculate heigMidpoint Estimation: Use the x-value in the middle of each rectangle to calculate heig

Calculate at each left, right or midpoint value to find the heights.Note: For increasing functions, a left-hand sum will give an under-estimate, and a right-han

sum will give an over-estimate. For decreasing functions, a left-hand sum will give an over-

estimate, and a right-hand sum will give an under-estimate.

Step 3: Find the areas of all the rectangles by multiplying width and height. Add these togethto estimate the area under the curve.

Left Riemann Sum Right Riemann Sum



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Introduction to Riemann Sums:Riemann sums is a method for approximating the total area underneath a curve on a graph, alsoknown as an integral.

This is a graph of the line = Suppose we want to find the area under the curve for the

interval = 2to = 8(starting at 2, ending at 8 along thex-axis)

We can split the graph into a finite number of rectangles, and use these to approximate the area.

As the rectanglesbecome smaller weget a better estimationof the area under thecurve. (The true areain this example is168.)

Area using 3 rectangles is 166. Area using 6 rectangles is 167.5.



= +cos Take first derivative = 3 s in Power Rule = 3 sin Take second derivative = 6 c o s Power and Trig RulePractice:Find the second and third derivatives of the following functions:

1. cos22. ln + 3. 2

Solutions:

1.

= cos2 Take first derivative

= sin2 2 Chain Rule

= 2sin2 Rewrite = 2sin2 Take second derivative =2cos2 2 Chain Rule =4cos2 Simplify = 4cos2 Take third derivative = 4 sin2 2 Chain Rule =8sin2

2. = ln + Take first derivative = + 3 Logarithmic and Power Rule



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= + 3 Take second derivative = + 6 Power Rule = +6 Take third derivative

= + 6 Power Rule3. = 2 Take first derivative

= 2 Exponential Rule = 2 Take second derivative = 2 Exponential Rule = 2 Take third derivative = 2 Exponential Rule

Critical Points and the Derivative Test

Knowing the first and second derivatives can reveal valuable information about a function. Mostimportantly, we can determine where the function is increasing, decreasing, concave up, andconcave down. Critical points are points where the function changes from increasing todecreasing (this is called a maximum), decreasing to increasing (this is a minimum), or changesin concavity (this is called a point of inflection).



b. lim 5. Find each limit at infinity using the lead terms of the polynomials.

a. lim b. lim c.

lim



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If the degree ofpis less than the degree of q, the limit is 0.

Examples: lim = lim = lim = 0

Let and be real numbers, and let be a positive integer.1. Scalar multiple: lim 2. Sum or difference:

3.

Product:

4. Power: lim[] =[lim ]

1. Find each limit using direct substitution.

a. lim + 3 + 1b. lim 27c.

lim 2. Find a right-hand and left-hand limit as approaches 2for the function = 3 5 2 2 + 4 > 2

Then determine whether the limit lim exists.3. Simplify each expression to find the limit.

a. lim b.

lim 4. Find each limit using LHopitals rule.a. lim



First Derivative Test

To find maximums and minimums, take the first derivative and set it equal to zero. Solve for These are your first critical points.

Example: = 3 4 3 = 9 49 4 = 0

= 49 = 23The first two critical points are

and . But we dont yet know whether these are minimumsmaximums. We need to examine the first derivative to figure that out. To do this, imagine orsketch a number line with the critical points labeled as shown in the figure below.



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Next, choose a point for each of the three ranges. We want to see if the derivative is positive ornegative in each range. This will tell us if the function is increasing (positive slope) or decreasing(negative slope).

From the above picture, you can see that -1, 0, and 1 will be good choices. Now, evaluate for each of the values x = -1 , x = 0 , and x = 1.

1 =91 41 = 9 41 = 5Positive (+)

0 =90 40 = 0 40 = 4

Negative (-)

1 =91 41 = 9 41 = 5Positive (+)

Range 1 Range 2 Range 3



When lim lim , we say that lim does not exist.If lim = lim , then lim exists.

, , The limit of a rational function where the numerator approaches some positive value and thedenominator approaches zero is infinity. If the numerator approaches a negative value, and th

denominator approaches zero, the limit is negative infinity.

Examples: lim = lim = Often you will be asked to take the limit of a rational function f(x) = p(x)/q(x) wherep(x) andq(x)bothapproach zero or infinity. Here are some tips for finding these limits:

Try to simplify the expression:

Example: lim = lim = lim = Take the derivative of the top and bottom, and use LHospitals rule:

Ifp(x), q(x) = 0 or

, then lim /=

lim /

Example: lim =l im = lim =

Often you will be asked to take the limit of a function as x approaches infinity.

When taking the limit at infinity of a rational function f(x) = p(x)/q(x) wherep(x) and q(x) arepolynomials:

lim /

If the degree ofpis greater than the degree of q, then the limit is positive or negativeinfinity depending on the signs of the leading coefficients;

If the degree ofpand qare equal, the limit is the leading coefficient of pdivided by thleading coefficient of q;



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We can also obtain this result graphically.

For discontinuous functions (such as some piecewise functions), the limit at a point may notexist. Instead, you can use a right-hand or left-hand limit.

Example:

= < 1 1 1

Here, the limit at 1 takes on two differentvalues, depending on whether you approach = 1from the left (smaller values) orfrom the right (larger values).

Right limit: lim = 0 Left limit: lim = 1

Although the function is discontinuous at

= 1, we can see that approaches 3when is close to 1.


2008 Rose-Hul

Evaluatingwe see that theThis tells us that the first critical

Here is a picture of this function:

Second Derivative Test

We can do the same things with

time the critical points will be poconcave up and a negative secon

above problem by taking the sec

Our critical point is x=0. Lets s

range.

Ra

increasing


slope of the graph is positive, then negative, an

point is a maximum, and the second critical poi

:

the second derivative that we did with the first d

ints of inflection, and a positive second derivatid derivative will indicate concave down. Lets c

nd derivative and setting it equal to zero.

= 9 4

= 18

18 = 0

= 0

t up a number line so we can choose a point in

maximum

minimum

ge 1 Range 2

increasing

decreasing

e

Page 17 o

d then positiv

nt is a minim

erivative. Th

ve will indicontinue the

ach importa



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We will use -1 for range 1, and 1 for range 2. Evaluate the second derivative at each of thesepoints.

=181 = 1 8 1

1 =18

Negative (-)

=181 = 1 8 11 = 18Positive (+)

This tells us that the function is concave down from to 0, and concave up from 0 to .Looking at the graph of the function, we can see this is correct.



The limit of a function at a value is the value that approaches as gets closer to lim

The limit as

approaches

of

lim = if and only if for every > 0 , there exists such that 0 < < imp < .

Some limits can be evaluated by simply substituting for in the function . Example: lim = 3 = 9

Direct substitution works anytime is continuous and can be calculated. This appliconstant functions, polynomial functions, and rational functions whenever is not infinity orzero.

However, limits are more often used when is discontinuous or is undefined.

A limit can often be estimated by making a chart of values of for values of whichapproach . If the values of approach a number, it is a good guess for the limit.

Example: lim

.9 .99 .999 1 1.001 1.01 1.1

2.710 2.970 2.997 3.003 3.030 3.310x-values approaching 1 from the left. x-values approaching 1 from the rig

f(x) approaches 3. f(x) approaches 3.



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Negative (-)

= cos = 1Positive (+)

This function is concave up on the interval , , concave down on the interval , and concave up on the interval , .

Check by graphing:



Example:Find the intervals of increasing, decreasing, concave up, and concave down for

= c o s On the interval

2,2.

First lets find the critical points for the first derivative:

= cos =sin s i n = 0 = 0 , , Now, set up a number line.

We will evaluate the first derivative at x =

,

,, and

.

=sin

ConcaveDown

ConcaveUpPoint of

Inflection

Range 1 Range 2 Range 3 Range 4



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32 = s i n 32 32 = 1Negative (-)

2 = s i n

2

2 = 1Positive (+) 2 = sin 2 2 = 1

Negative (-)

32 = s i n 32 32 = 1Positive (+)

is a minimum, 0 is a maximum, and is a minimumThis function is increasing on the interval ,0 ,2, and decreasing on the interval2, 0, .

Check by graphing:



Now, lets do the second derivative test. To do this, evaluate the second derivative and set it eqto zero to find the critical points.

= sin = cos cos = 0

The critical points are x = , , , .

We will evaluate the second derivative at ,0,and . = cos = cos

= 1

Positive (+)0 = cos 00 = 1


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