Terrio's Reflection

Aleks blog...so, finding the slope of a line when the equation is given to you:

Notes - Take the equation given, turn into slope intercept for, pull out the slope from equation, that's the answer.

Example: find the slop of the line 4 x +5 y = 3

You start out by converting the equation to slope intercept form, which is y = m x + b. ( m stands for the slope )

4 x + 5 y = 3
5 y = 3 - 4 x
y = ( 3 - 4 x ) / 5
y = ( 3 / 5 ) - ( 4 x / 5 )
y = - ( 4 /5 ) x + ( 3 / 5 )

Now that it is converted into slope intercept form you take ( m ) out of the equation, - ( 4 /5 ), and that's your slope.

So the answer is - ( 4 /5 )

Reflection

In calculus this past week we did stuff with substitution, so you need to know how to integrate and find bounds for some substitution problems. Lately we have been doing a lot of stuff with integrals, so we definitely need to know how to integrate. Here is an example of integration.

EXAMPLE:

1. Find the integral of: x^3 + 3x
When finding the integral, you have to add one to the exponent and divide that number.
x^4/4 + 3(x^2/2)
1/4x^4 + 3/2x^2 + c

2. Find the integral of: 6x
6[x^2/2]
3x^2 + c
Also when solving for integrals, do not forget to add the +c.

When finding the bounds, you set the equations in the problem equal too each other and then solve for x.

Ex :

x^2 + 2x = 3x

x^2 - x = 0

x ( x - 1 ) = 0

the bounds are x = 1, 0

blog march 24

Well i just finished 28 aleks and it was very sick nasty, this stuff is so simple but the simple mistakes are starting to kill me. Its driving me insane, but at least i am done and will be getting my 28 points, but ill review some things i got wrong
Factoring a quadratic with a leading coefficient greater than one...
Ex :
5y^2 + 13y - 6
Use the FOIL method backwards...
The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )
Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13
You'll find that 3 and -2 work...3 in place of p and -2 in place of q.
Factoring a quadratic with a leading coefficient greater than one...
Ex :
5y^2 + 13y - 6
Use the FOIL method backwards...
The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )

Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13

You'll find that 3 and -2 work...3 in place of p and -2 in place of q.

The answer is

Terrio's Reflection

Don't really understand what we're doing this week too well sooooo, review blog? I'll review chain rule, veritcal asymptotes, and derivatives:

-When solving for a vertical asymptote of a graph, you factor out the top and bottom and set the bottom equal to zero.

Examples:
1) f(x) = x^2/x^2-4
= x(x)/(x+2)(x-2)
(x+2)(x-2) = 0
x = -2, 2

2) g(x) = 1/2(x+1)
= 2(x+1) = 0
2x+2 = 0
x = -1

-When solving chain rule, you take the derivative from outside in: derivative of outside, recopy inside, multiply by derivative of inside.

Examples:
1) y = cos3x^2
y' = -sin3x^2 X (6x)
= -6xsin(3x^2)

2) g(x) = 3(4-9x)^4
= 12(4-9x)^3 X (-9)
= -108(4-9x)^3

-When taking regular derivatives you can use a few different strategies, sometimes you may have to use the product rule or quotient rule.

Examples:
1) f(x) = 3x^5 + 4x +8
f'(x) = 3(5)x^(5-1) + 4 +0
= 15x^4 + 4

2) f(x) = 3xy + 4x^2
f'(x) = [3x(1) + y(3)] + 8x
= 3x + 3y + 8x
= 11x + 3y

3/20 Reflection

Last week we learned something new, Integration by Substitution. This is, in a way, the product/quotient rule of integration because it allows you to integrate terms that are being multiplied and divided.

To use substitution, one function must be the derivative of another.
*If you are only off by a number, tho, it can be fixed.

Here is an example of substitution:

x ( x^2 + 1 )^3

-First, find your "u" and your "du"

u = x^2 + 1 du = 2x

The problem has to become 1/2 S 2x ( x^2 + 1 )^3 because:
du = 2x (du is the derivative of u) that means that on the front side of the integration symbol, you must put the reciprical of 2. This is to balance the eqn out because you had to add a 2 to the problem.

-Now replace the terms with u and du:
1/2 S u^3 du

-Now integrate:
1/2 (1/4u^4) + c

-Now plug in for u and simplify and you get:
1/8 (x^2 + 1)^4 + c

BLOG MARCH 20

This week in cal we did some integrating thing, which im having a little difficulty with but i am slowly starting to get it. I didnt bring my notes home on it, so ill talk about some aleks stuff.
Factoring a quadratic with a leading coefficient greater than one...
Ex :
5y^2 + 13y - 6
Use the FOIL method backwards...
The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )
Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13
You'll find that 3 and -2 work...3 in place of p and -2 in place of q.
Tomorrow we have a quiz on the integration stuff, i hope i do good.

Reflection

So I forgot to bring my notes home but I will attempt to this blog and get it close to being right. This week in calculus we did more Aleks, and on there we are reviewing simple things which we have learned before like dealing with negative exponents and even factoring…basically easy. But we began to learn a new section in class which has to deal with substitution. When substituting integrals, you have to first find the u and du. The du is the derivative of u, but there is an exception where if they are one number off then you can still use those numbers; if it is off by more than one number then you have to do the reciprocal of a common number. During the substitution process, you have to use the process of integration also. By the end of the problem, something out of the problem will disappear and it will end with +c.

mardi gras blog

This is a quick review on thingsa we did on the exam and stuff we did before the break, i hope everyone enjoyed themselves.
Example: Find an equation of the tangent line to the graph of f at the given point
f(x) = x/x+4, (-5, 5)
= (x+4)(1) -[(x)(1)] / (x+4)^2
= x+4-x / (x+4)^2
= 4/(x+4)^2
m = 4/(-5+4)^2 = 4/1^2 = 4
point slope: y - 5 = 4(x + 5)
Example for a derivative:
use shortcut
f(x) = x^2
= 2x
Quotient Rule:
f(x)/g(x) = g(x)f’(x) – [f(x)g’(x)] / (g(x))^2
(recopy bottom)(derivative top) – [(recopy top)(derivative bottom)] / (bottom)^2
Example: f(x) = x / x^2 + 1
f’(x) = (x^2 + 1)(1) – [(x)(2x)] / (x^2+1)^2
= x^2 + 1 – 2x^2 / (x^2 + 1)^2
= -x^2 + 1 / (x^2 + 1)^2
STEPS to solving implicit derivatives:
1. Differentiate both sides with respect to x, d_/dx
2. Collect all dy/dx terms on one side, and move the other terms to the other side.
3. Factor out dy/dx.
4. Solve for dy/dx.
5. Simplify, *replace with original equation if possible.**

Lindsey's Reflection

well since we did not have
school this week we did
not learn anything neew.
so once again i will have to
do a blog on something we
previously learned.

So i am gonna do my
blog on the product rule


Product Rule:

So for the product rule
you recopy the first
and multiply it by the
derivative of the first.
then you recopy the
second and multiply it
by the derivative of the first.
then you add these two
together.

Here is an example:

(3x^4)(6x)

-3x^4(6)+6x(12x^3)

the final answer to this problem
is gonna be

18x^4+72x^4
=90x^4

See evevryone tomorrow!

End of Mardi Gras Reflection

Ok so i needed something old to review for this weeks blog an I didn't wanna repeat somethin so here's one I never did: Disks an Washers!

There are 2 formulas you have to know:

Disk Formula: ( pi )S( r^2 )dx
Washer Formula: ( pi )S( top^2 )-( bottom^2 )


EXAMPLE: Find the volume of the solid formed by revolving the region bounded by the
graphs of y = (square root of (x)) and y = x^2 about the x-axis.
First, you plug in the 2 equations into your calculator and graph them.


For this graph, its a washer, an we know that from graphin it in our calculadoras.
-To find the bounds, set the 2 eqns = to each other
(square root of (x))^2 = (x^2)^2
x = x^4
x^4 - x = 0
x(x^3 -1) = 0
x = 0, 1
Plug into the formula.
(pi)1/S/0 (square root of (x))^2 - (x^2)^2
(pi)1/S/0 x - x^4
Find the integral.
pi[1/2x^2 - 1/5x^5]
Plug in the bounds.
(pi)[1/2(1)^2 - 1/5(1)^2] - [1/2(0)^2 - 1/5(0)^2]
= 3pi/10

Reflection

To review some things which we learned this year and which keep reapplying to more sections in calculus.

To find an equation of a tangent line, you must take the derivative of the function and plug in the x-value. (If not given a y-value, plug into the original equation to get the y-value.) Then plug the values into point slope which is: y - y1 = m(x - x1)
Example: Find an equation of the tangent line to the graph of f at the given point
f(x) = x/x+4, (-5, 5)
= (x+4)(1) -[(x)(1)] / (x+4)^2
= x+4-x / (x+4)^2
= 4/(x+4)^2
m = 4/(-5+4)^2 = 4/1^2 = 4
point slope: y - 5 = 4(x + 5)

To find a regular derivative you can use the whole limit process or the shortcut.
Example:
use shortcut
f(x) = x^2
= 2x


There are a certain set of guidelines to follow when solving related rate problems:
1. Identify all given quantities and quantities
to be determined. Make a sketch and label the
quantities.
2. Write an equation involving the variables
whose rates of change either are given or are
to be determined.
3. Using the Chain Rule, implicity differentiate
both sides of the equation with respect to
time t.
4. After completing Step 3, substitute into the
resulting equation all known values for the
variables and their rates of change. Then
solve for the required rate of change.

Example:
xy = 4
Step 1: dy/dt = ? x = 8 dx/dt = 10
Step 2: xy = 4
Step 3: xdy/dt + ydx/dt = 0
Step 4: xdy/dt = -10y
= -10y/x

Terrio's Reflection

Okay so I did all fifty Aleks' for the week and my head is about to explode. I don't think I have ever done so much math in three days in my life so I'll do my blog on one of the Aleks modules. Factoring a quadratic with a leading coefficient greater than one...

Ex :

5y^2 + 13y - 6

Use the FOIL method backwards...

The coefficient of is . Using whole numbers, only 1 and 5 can be multiplied to get .
So you have to find two integers and such that 5y^2 + 13y - 6 = ( 1y + p) ( 5y + q )

Using the FOIL method to expand the right-hand side of this equation, you get that ( pq ) = -6 and q + 5p = 13

You'll find that 3 and -2 work...3 in place of p and -2 in place of q.

The answer is

blog

well right now im doing some aleks and they are starting to get on my nerves bc i keep making simple mistakes and have to keep reworking the problems 412378942175378 times, and im about to kill some math. Here are examples of why im about to murder this website

Adding rational expressions with common denominators
Subtract. Write your answer in lowest terms.

Each fraction has the same denominator, namely .
So we subtract the numerators and keep the denominator the same to get
.
Distributing the negative signs over the parentheses, we have
.
Combining like terms, we get
.
This fraction cannot be simplified.
The answer is .

Lindsey's Reflection

This past week was our exams.

it wasnt as bad as i thought but you
never know with calculus.

Anyway for this blog i am gonna
review rolles theorem, mean value
theorem, and the first derivative test.

Rolles Theorem:

first and foremost you
have to make sure that
it is continuous and
differentiable.
if it is neither you can not
apply the theorem to whatever
problem is given.

if it it contionuous and
differentiable then you have
to make sure that the two y values
match.
and once again if they do not
match the theorem can not
be applied.

if it applies you then take the derivative
set it equal to zero, and solve for x.

this theorem is used to find a max or
a min.

Mean value theorem:

with this theorem you also
have to make sure it is continuous
and differentiable.

you then take the derivative and set it equal to
the slope and solve.

First derivative Test:

1. take the derivative and
set it equal to zero. Solve for x.

2.find where it is not
differentiable.

3. set up intervals using the
numbers you found in steps
1 and 2.

4. plug in a number from
each interval into the original
equation to find if it is
negative or positive.

5. if it is positive it is increasing
if it is negative it is decreasing.

6. determine the max or min from
step 5.

Not too hard.

i hope that everyone has a good mardi gras.

Reflection

This past week we reviewed derivative, absolute extrema, critical numbers, etc. for our 3rd nine week exam. THis is just to review the stuff we have learned.
Example: Find an equation of the tangent line to the graph of f at the given point
f(x) = x/x+4, (-5, 5)
= (x+4)(1) -[(x)(1)] / (x+4)^2
= x+4-x / (x+4)^2
= 4/(x+4)^2
m = 4/(-5+4)^2 = 4/1^2 = 4
point slope: y - 5 = 4(x + 5)

Example for a derivative:
use shortcut
f(x) = x^2
= 2x

Quotient Rule:
f(x)/g(x) = g(x)f’(x) – [f(x)g’(x)] / (g(x))^2
(recopy bottom)(derivative top) – [(recopy top)(derivative bottom)] / (bottom)^2
Example: f(x) = x / x^2 + 1
f’(x) = (x^2 + 1)(1) – [(x)(2x)] / (x^2+1)^2
= x^2 + 1 – 2x^2 / (x^2 + 1)^2
= -x^2 + 1 / (x^2 + 1)^2

STEPS to solving implicit derivatives:
1. Differentiate both sides with respect to x, d_/dx
2. Collect all dy/dx terms on one side, and move the other terms to the other side.
3. Factor out dy/dx.
4. Solve for dy/dx.
5. Simplify, *replace with original equation if possible.**

Word Problems
+ Example:
Air is being pumped into a spherical balloon at a rate of 4.5 cubic feet per
minute. Find the rate of change of the radius when the radius is 2 ft.
(*word problem taken from notes)
1. dv/dt = 4.5 ft^3/min dr/dt = ? r = 2 ft.
2. V = 4/3pir^3
3. dv/dt = 4/3pi[3r^2dr/dt]
= 4pir^2dr/dt
4. 4.5 = 4pi(2)^2dr/dt
4.5 = 16pidr/dt
dr/dt = 9/32pi ft/min.

Lindsey's reflection

Im gonna do this blog
on the different types
of discontinuities.

the first one is a removable.
A removable is when the
graph is not defined at
a specific point. at
a removable the limit
exists and the function is
continuous everywhere
except at that point.

the second one is a jump.
this is when there are two
different points on the graph.
the limit does not exist and it is
continuous everywhere
except at the jump.

An infinate is next. it is
also known as an
asymptote. you can
either have vertical asymptotes
or horizontal asymptotes. the limit
may or may not exist and the function
is continuous everywhere except at
the asymptote.

the fourth and final is oscillation.
this would be an extreme oscillating
graph. the limit does not exist
and the function is not continuous.

Goodnight!

Terrio's Reflection

This past week we reviewed a lot on derivatives and we will have a lot of them on the Exam this coming week so I'll blog on Product and Quotient Rule to review the two...

-Product Rule :
f( x ) g( x ) = f( x ) g’( x ) + g( x ) f’( x ) ( copy 1st ) ( derivative 2nd ) + ( copy 2nd ) ( derivative 1st )

Ex. g( x ) = ( 3x^2+4 ) ( 2x^2-3x )
= ( 3x^2+4 ) ( 4x-3 ) - ( 2x^2-3x ) ( 6x )
= 12x^3 - 9x^2 + 16x - 12 - 12x^3 + 9x^2
= 16x - 2

-Quotient Rule :
f (x ) / g( x ) = g( x ) f’( x ) – [ f( x ) g’( x ) ] / ( g( x ) )^2 ( copy bottom ) ( derivative top ) – [ ( copy top ) ( derivative bottom ) ] / ( bottom )^2

You solve the same way with the quotient rule, following the steps ( formula ) and when simplifying, if there is a trig identity in the answer use the identity to simplify the answer completely.

Reflection

For our 3rd nine weeks exam, we are reviewing chapters 2 and 3 to go over how to solve different types of derivative equations. An explicit function is solved for y, and an implicit is not solved for y. When you take a derivative you must put the notation dy/dx by any variable except for x.
STEPS:
1) Differentiate both sides with respect to x. d_/dx
2) Collect all dy/dx terms on one side, and move the other terms to the other side.
3) Factor out dy/dx.
4) Solve for dy/dx.
5) Simplify, *replace with original equation if possible.**

Examples:

1) x^2 + y^2 = 9
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
= -x/y

2) x^3 + y^3 = 64
3x^2 + 3y^2dy/dx = 0
3y^2dy/dx = -3x^2
dy/dx = -3x^2/3y^2
= -x^2/y^2

Section 2.2
Example:
Find the average rate of change of the function over the given interval.
f(t) = 4t +5 [1, 2]
(1, 9) = 4(1) + 5 = 9
(2, 13) = 4(2) + 5 = 13

m = y2 – y1 / x2 – x1
= 13 – 9 / 2 -1
= 4/1
=4

blog before exam

in calculus we have been reviewing over alot of old material preparing for our third weeks exam. I hope i do well on it. We are aslo still working on something called an ALEKS and its helping me rememeber a lot of old stuff that i might of forgotten. In calc though i will go over the product and quotient rule bc it will be on our exam.
Product Rule: ( copy 1st ) ( derivative of 2nd ) + ( copy 2nd ) ( derivative of 1st )
Quotient Rule: ( copy bottom ) ( derivative top ) – [( copy top ) ( derivative bottom )] / ( bottom ) ^2
-Product Rule Ex :
g (x) = (3x^2+4) f (x) = (2x^2-3x)
= ( 3x^2+4 ) ( 4x-3 ) - ( 2x^2-3x ) ( 6x )
= 12x^3 - 9x^2 + 16x - 12 - 12x^3 + 9x^2
= 16x – 2
-Quotient Rule Ex :
x– 2 / x^2 + 2
( x^2 + 2 ) ( 1 ) – ( x – 2 ) ( 2x ) / ( x^2 + 2 )^2
-x^2 – 4x + 2 / (x^2 + 2 )^2

Lindsey's Reflection

This week we had a big review
on derivatives.
we just got done
learning about intergrals and
since it is the antiderivative
we needed this review.

Here is just a brush up of derivatives:

3x^6+4

in this case you would just
bring the 6 to the front and multiply it
by 3.
then you would subtract one
from the exponent.

this would leave you with

18x^5

this is the answer.
when you take the derivative
of 4 it is going to be zero
because there is no variable,

that is the basic type of derivative.

we also reviewed other types such '
as the product rule, quotient rule
and the chain rule.

see everyone tomorrow. goodnight!

Terrio's Reflection

Soooooooo for some strange reason this ALEKS thing is not working for me. It says my username or password is incorrect. We did some review stuff this week on quotient rule and product rule so I'll recap these two formulas a little bit and use an example for each.

Product Rule: ( copy 1st ) ( derivative of 2nd ) + ( copy 2nd ) ( derivative of 1st )

Quotient Rule: ( copy bottom ) ( derivative top ) – [( copy top ) ( derivative bottom )] / ( bottom ) ^2

-Product Rule Ex :
g (x) = (3x^2+4) f (x) = (2x^2-3x)
= ( 3x^2+4 ) ( 4x-3 ) - ( 2x^2-3x ) ( 6x )
= 12x^3 - 9x^2 + 16x - 12 - 12x^3 + 9x^2
= 16x – 2

-Quotient Rule Ex :

x– 2 / x^2 + 2
( x^2 + 2 ) ( 1 ) – ( x – 2 ) ( 2x ) / ( x^2 + 2 )^2
-x^2 – 4x + 2 / (x^2 + 2 )^2

blog feb 20,2011

This week in calculus we been reviewing from some old stuff we did in the earlier part of the year. We had a quiz and hopefully i did pretty well on it. We also have been doing something called an ALEKs. I think i'm going to like doing this stuff bc it made me realize i forgot hot to do some simple problems. I aslo went visit ULL this weekend and the engineer teacher told us how important it is that we remember this simple math. I'm looking forward to reteaching myself this stuff. Anyways bc we did old stuff in calc i will bring up some old formulas. You guessed it, the famous quotient and product rule formula. This week we mostly learned about how to apply the product rule and the quotient rule.
here is the formula for the product rule:
(recopy the first equation)(take the derivative of the second equation)+(recopy the second equation)(take the derivative of the first equation)
here is a example where you would have to
apply this rule:

(3x-2x^2)(5+4x)
since these two equations are being
multiplied you have to use the product
rule and plug in the formula
After doing so this is what you end up with
3x-2x^2)(4)+(5+4x)(3-4x)
then you simplify as much as
you can
12x-8x^2+15-20x+12x-16x^2
=-32x^2+4x+15
next is the quotient rule
here is the formula:
(bottom)(derivative of top)-[(top)(derivative of bottom)]/ (bottom)^2
the only way that this rule would not apply is if
you have a number instead of an equation
at the bottom of the fraction.
anyway here is an example using
the quotient rule:
5x^2-2/x^2+1
(x^2+1)(5)-[(5x-2)(2x)/(x^2+1)^2
from here, once again, you just
simplify as much as possible
5x^2+5-[10x^2-4x]/(x^2+1)^2
always keep the bottom of the problem
as is.
5x^2+5-10x^2+4/(x^2+1)^2
=-5x^2+4x+5/(x^2+1)

Reflection

This week we reviewed information from chapter 2.
Chain Rule :
you take the derivative of the outside, then derivative of the outside then recopy the inside, multiply by derivative inside.

Derivative of a natural log:

d/dx lnu= 1/u x du

examples:

d/dx ln2x= 1/2x x 2= 1/x

d/dx ln(x^2+1)=1/x^2+1 x 2x= 2x/ x^2+1

d/dx xlnx= x(1/x x 1) +lnx(1)=1+lnx

d/dx lnx^3= 1/x^3 x 3x^2= 3/x

d/dx (lnx)^3= 3(lnx)^2 x 1/x x 1=

3(lnx)^2/x

Implicit functions are not solved for y . Explicit functions are solved for y . When you take the derivative of y you have to put either dy/dx or y' after every y . For x you just take the derivative like normal and constants are always 0 . When solving all you do is take the derivative of the whole equation . After you take the derivative you solve for dy/dx or y ' . And then simplify your answer.

Related Rates
+ When solving related rates problems, you must follow the guidelines.
+ Examples:
1) y = squareroot(x)
1. dy/dt = ? x = 4 dx/dt = 3
2. y = squareroot(x)
3. dy/dt = 1/2x^-1/2
4. dy/dt = 1/2(4)^-1/2
= 3/4
2) xy = 4
1. dy/dt = ? x = 8 dx/dt = 10
2. xy = 4
3. xdy/dt + ydx/dt = 0
4. xdy/dt = -10y
= -10y/x

Blog number is unknown

Since all our problems have integrating in it, i will talk about integration as a review. All integration is is doin the opposite of takin the derivative. Instead of multiplyin the coefficient by the exponent then subtractin one from the exponent like you do when you take the derivative, you ADD one to the exponent and DIVIDE the coefficient by the exponent. Another thing is for these section you need to know how to graph the equations and
how to find there reflections on a graph.

We learned how to solve for volume, which you have some different formulas to use.
One of those formulas is (pi)S(r^2)dx
Another formula is (pi)S(top^2)-(bottom^2)dx
When you have any problem where you are solving for volume, you need to square the
equation first, then you would integrate like usual and plug in the bounds. But while doing this keep the pi out front and do not add it until the end.

reflection 2/13

Ok so something basic because i'm tired and i want to go to sleep. Integration has become the new derivative in calculus (even tho technically its the anti derivative) Everything we have been doing lately has somethin to do with integrals, so you HAVE TO KNOW HOW TO INTEGRATE. Here are some examples of integration.

EXAMPLE:

1. Find the integral of: x^3 + 3x

x^4/4 + 3[x^2/2]

1/4x^4 + 3/2x^2 + c
*You have to add the "+ c" to the end of your problem after you have integrated or it is NOT right.

2. Find the integral of: 6x

6[x^2/2]

3x^2 + c

-Its pretty simple, so make sure you know it! And if you dont know it, LEARN IT! Or you WILL fail.

Terrio's Reflection

Two things we NEEDDDDDDDDDDDDDDD to know for Calc right now are Integration and how to determine the bounds of a region on a graph. Both are fairly easy to do. For Integration all you do is add one to the exponent and then divide the new exponent by the number out in front of the variable. You can't forget to put a c at the end of the equation. c stands for the constant at the end of a problem because when working a derivative backwards you never know if there was a constant in the problem or not.

Ex :

2x
-Add one to the exponent and get 2 x ^ 2
-Divide the front number by the exponent and you get ( 2/2 ) x ^ 2
The answer is x ^ 2 + c

-For finding the bounds you set the equations in the problem equal too each other and then solve for x.

Ex :

x^2 + 2x = 3x

x^2 - x = 0

x ( x - 1 ) = 0

the bounds are x = 1, 0

Reflection

To begin, you must know how to solve an integral to work any of the problems that
we are working on in class right now.
1. Find the integral of: x^3 + 3x
When finding the integral, you have to add one to the exponent and divide that number.
x^4/4 + 3(x^2/2)
1/4x^4 + 3/2x^2 + c

2. Find the integral of: 6x
6[x^2/2]
3x^2 + c
Also when solving for integrals, do not forget to add the +c.

Another thing is for these section you need to know how to graph the equations and
how to find there reflections on a graph.

We learned how to solve for volume, which you have some different formulas to use.
One of those formulas is (pi)S(r^2)dx
Another formula is (pi)S(top^2)-(bottom^2)dx
When you have any problem where you are solving for volume, you need to square the
equation first, then you would integrate like usual and plug in the bounds. But while doing this keep the pi out front and do not add it until the end. Most of this is simple, you just have to make sure you know the steps and solve all the problems correctly.

Blog ??????

Since all our problems have integrating in it, i will talk about integration as a review. All integration is is doin the opposite of takin the derivative. Instead of multiplyin the coefficient by the exponent then subtractin one from the exponent like you do when you take the derivative, you ADD one to the exponent and DIVIDE the coefficient by the exponent. A cool thing about integration versus derivatives is that there are no product rules, quotient rules, or chain rules which helps it to be a little simpler. Its a process jus like takin derivatives in that it is very easy if you practice it.

Here is an example on how to integrate:

3x^2
add one to the exponent: 3x^3
now divide the coefficient by the exponent ( or multiply by the reciprical ): 3/3x^3
and you get x^3

**a good way to check your answer is to take the derivative of your answer, and you should get your problem: the derivative of x^3 is 3x^2

Super Bowl Blog

wellll time for some bloggin now that the superbowl's over. I'll do a blog on integration because i think its the concept that i grasped best in calculus this whole year. I feel like for once calculus isn't so stressful! Ok so here's a few examples of integration:

ex. 1) 3x^2 to integrate, add one to the exponent and divide the coefficient by that exponent, so.... 3/3x^3 = x^3.......see, that one was simple.

ex. 2) x^3 okay, same thing, add one to the exponent making it ^4 then the divide the coefficient ( 1 ) by 4, which gives you 1/4x^4

ex. 3) ( x - 1 )^2 okay for this one you need to foil it out first before you can integrate: x^2 - 2x + 1....now integrate one term at a time:
x^2 = 1/3x^3 -2x = -x^2 and 1 = x
-now put it together and your integral for this example is 1/3x^3 - x^2 + 1

**HINT** If you want to check your answer, its simple, just take the derivative of the answer you get, and if you integrated properly, the derivative of your answer should equal the problem.

Terrio's Reflection

Integration seems like something important this quarter, we do it in just about everything we learned so far so I'll recap that right quick.

-Integration is similar to taking a derivative. What you do is you underive a derivative. Basically work a derivative backwards. You add one to the exponent and then divide the new exponent by the number out in front of the variable. One thing you HAVEEEEEE to remember is the c. c stands for the constant at the end of a problem because when working a derivative backwards you never know if there was a constant in the problem or not.

Ex:

2x
-Add one to the exponent and get 2 x ^ 2
-Divide the front number by the exponent and you get ( 2/2 ) x ^ 2
-After integrating the problem you will get x ^ 2 + c for your answer.

Reflection

In class, we have been learning about solving for volume after graphing a washer or a disk. Volume > S area of cross section.
Disk Method = (pi)S(r^2)dx
Washer Method = (pi)S(top^2)-(bottom^2)
When you revolve a graph, then you spin it around on the x-axis.

EXAMPLE: Find the volume of the solid formed by revolving the region bounded by the
graphs of y = (squareroot of x) and y = x^2 about the x-axis.
First, you plug in the 2 equations into your calculator and graph them.
After you have graphed then you draw the graph reflection to determine if it
is a washer or a disk. For this particular graph, it is a washer.
Set the 2 equations equal to find the bounds.
(squareroot of x)^2 = (x^2)^2
x = x^4
x^4 - x = 0
x(x^3 -1) = 0
x = 0, 1
Plug into formula.
(pi)1/S/0 (squareroot of x)^2 - (x^2)^2
(pi)1/S/0 x - x^4
Find the integral.
pi[1/2x^2 - 1/5x^5]
Plug in the bounds.
(pi)[1/2(1)^2 - 1/5(1)^2] - [1/2(0)^2 - 1/5(0)^2]
= 3pi/10

EXAMPLE: Find the volume of the solid formed by y = 4 - x^2 revoved about the x-axis
in Quadrant I.
First graph the equation.
(pi)2/S/0(4-x^2)dx
(pi)2/S/0(16-2x^2+x^4
Find the integral and plug in bounds.
(pi)[16(2)-2/3(2)^3+1/5(2)^5] - [16(0)-2/3(0)^3+1/5(0)^5]
= 256pi/15

prompt unknow number

I’m going to do a throw back as well.
When doing limits at infinity you do: 1. if the degree of the top=0 bottom
you divide the coefficients.

2. if the top of the degree is greater it equals
positive or negative infinity. plug in large value
to see if it is negative or positive.

3. if the top degree is less than the bottom
degree it equals zero

if it is not a fraction use a table. plug in 100,
1000,10000 until you see a pattern.
here is an example:
lim 2x+5/3x^2+1= 0
x-->infinity
the answer to this problem is zero because
the top degree is less than that of the
bottom degree! function is contionuous or there is a
discontinuity.
-a function is continuous is it does not contain
one of the four types of discontinuities.
-to be continuous a function must have a limit
at every point on the interior and be defined at
the limit of the point.

here are the four types of discontinuities to
look for:

1. removable- when the graph is not defined at a
point. (open circle).
- the limit exists

- the function is conti uous everywhere except
at that point. therefore, if we are talking about
the function as a whole we say that it is not
continuous.

2. jump
- the limit does not exist

- the function is continuous everywhere except
at the jump. it is not continuous as a whole.

3. infinate- an asymptote
- the limit may or may not exist

- the function is continuous everywhere except
at the asymptote. it is not continuous as a
whole.

4. oscillation- an extreme oscillating graph
- the limit does not exist

- the function is not continuous.

Reflection 1/30

We are learning integration and TRAM, LRAM, RRAM, and MRAM in calculus right now. It is pretty easy if you pay attention in class and do your homework. I am just having problems with the TRAM and MRAM. I need more practice on those types of problems. But I can show some examples on integrations because those are the easiest.

EXAMPLE:

1. Find the integral of: x^3 + 3x

x^4/4 + 3[x^2/2]

1/4x^4 + 3/2x^2 + c
*You have to add the "+c" to the end of your problem, after you have integrated. And that is your answer. Pretty simple, huh?

2. Find the integral of: 6x

6[x^2/2]

3x^2 + c

-See, this stuff is actually pretty easy. I hope that this stuff does not get any harder though, because I'm actually doing good right now.

Terrio's Reflection

Throwback time. I'll review something that I randomly flip to in my binder today....

-Implicit functions :

Implicit functions are not solved for y .

-Explicit functions :

Explicit functions are solved for y .

-Taking the derivative:

When you take the derivative of y you have to put either dy / dx or y' after every y . For x you just take the derivative like normal and constants are always 0 . When solving all you do is take the derivative of the whole equation . After you take the derivative you solve for dy / dx or y ' . Last you simplify your answer if possible .

Ex :

3 x+2 y = 6

= 3 x+2 y dy / dx = 0

= 2 y dy / dx = -3 x

= dy / dx = - ( 3 x / 2 y )

The answer is dy / dx = - ( 3 x / 2 y ) .

Lindsey's Reflection

This week i am going to go
back and do this blog on
limits at infinity just to brush
up on this concept.

---if it is a fraction there are three
rules that you have to use when
trying to solve it.

1. if the degree of the top and bottom are
the same you divide the coefficients.

2. if the top degree is greater than the
bottom degreee it is going to equal
either infinity or negative infinity.
you have to plug in large values
in your table to see which one of these
it is going to be.

3. if the top degree is less than the bottom
degree it is going to equal zero.

---if it is not a fraction use a table.
you plug in 100,1000,10000 and so
on until you see a pattern between the
numbers.

here are some examples using these
rules:

lim 2x+5/3x^2+1 =0
x-->infinity

the answer is gonna be zero because the
top degree is less than the bottom degree.

lim 2x^2+3/4x^2+5=1/2
x-->infinity


the answer is gonna be 1/2 because the degrees
are the same so you would divide the coefficients.

see everyone tomorrow.

Reflection

Although we learned plenty of new stuff, I’m going to review some of the old stuff so I don’t forget how to work problems like these. Because with some of the new stuff, you have to understand the old information to work the problems.

Finding derivatives of trig.
1. f(x) = 3cosx - sinx/4
= -3sinx - (4)(cosx) - [(sinx)(0)] / (4)^2
= -3sinx-(4cosx/(4)^2)

2. y = 3x^2secx
= 3x^2secxtanx + (3(2))xsecx
= 3x^2secxtanx + 6xsecx

3. y = 1/2csc2x at (pi/4, 1/2)
= -csc2xcot2x
=> -csc2(pi/4)cot2(pi/4)
= 0rivatives with trig.
Related Rates
+ When solving related rates problems, you must follow the guidelines.
+ Examples:
1) y = squareroot(x)
1. dy/dt = ? x = 4 dx/dt = 3
2. y = squareroot(x)
3. dy/dt = 1/2x^-1/2
4. dy/dt = 1/2(4)^-1/2
= 3/4
2) xy = 4
1. dy/dt = ? x = 8 dx/dt = 10
2. xy = 4
3. xdy/dt + ydx/dt = 0
4. xdy/dt = -10y
= -10y/x
Here are the steps you need to solve for an implicit derivative:
1) Differentiate both sides with respect to x. d_/dx
2) Collect all dy/dx terms on one side, and move the other terms to the other side.
3) Factor out dy/dx.
4) Solve for dy/dx.
5) Simplify, and replace with original equation if possible
Example: x^3 + y^3 = 64
3x^2 + 3y^2dy/dx = 0
3y^2dy/dx = -3x^2
dy/dx = -3x^2/3y^2
= -x^2/y^2

Blog 1/23/11

We learned about integration and integrals, and im struggling a little with it but I hope I can manage the quiz tomorrow better than the hw I did today. All integration is is doin the opposite of takin the derivative. Instead of multiplyin the coefficient by the exponent then subtractin one from the exponent like you do when you take the derivative, you ADD one to the exponent and DIVIDE the coefficient by the exponent. Sx^4dx
= x^5/5
= 1/5x^5+c
you must always remember to put plus c
at the end of the answer or it would be
considered wrong.
This can also be called the anti derivative

Reflection 1/23

This week we learned all about integration. All integration is is doin the opposite of takin the derivative. Instead of multiplyin the coefficient by the exponent then subtractin one from the exponent like you do when you take the derivative, you ADD one to the exponent and DIVIDE the coefficient by the exponent. A cool thing about integration versus derivatives is that there are no product rules, quotient rules, or chain rules which helps it to be a little simpler. Its a process jus like takin derivatives in that it is very easy if you practice it.

Here is an example on how to integrate:

3x^2
add one to the exponent: 3x^3
now divide the coefficient by the exponent ( or multiply by the reciprical ): 3/3x^3
and you get x^3

**a good way to check your answer is to take the derivative of your answer, and you should get your problem: the derivative of x^3 is 3x^2

Lindsey's Reflection

This week we learned about integration.

here are a few examples of it:

Sx^4dx
= x^5/5
= 1/5x^5+c

you must always remember to put plus c
at the end of the answer or it would be
considered wrong. the answer to this
problem may also be referred to as the
anti derivative because it is the reverse
of taking the derivative.

here are more examples:

find the antiderivative of 3x.

S3xdx
=3[x^2/2]+c
=3/2x^2+c


S1/x^3dx
=Sx^-3dx
=x^-2/-2
=-1/2x^-2+c

Ssec^2+cosxdx
=tanx+sinx+c

Terrio's Reflection

This week we learned a new thing in Calculus. We learned how to do Integrations.

Basically all Integrating is, is reversing the steps of taking a derivative. You just work the problem backwards so instead of taking a derivative you're actually undoing a derivative.

Ex. :

The cubed root of 6x

that would be rewritten as 6 x ^ ( 1/3 )

Reverse the derivative = 6 x ^ ( 4/3 )

Divide the exponent by the number in front of the variable = ( 6/( 4/3 ) x ^ ( 4/3 )

Simplify this and you get ( 9/2 ) x ^ ( 4/3 )

Reflection

In calculus this week, we started learning information from chapter 4. We learned about integration and integrals.
To take a derivative you would:
F(x) = 2x^2
F’(x) = 4x
Now you are doing the opposite of taking the derivative.

For integration the symbol I’ll be using is S
Examples:
1. S(2x-3x^2)dx = S2x^2-3x^3dx
=2x^2/2 – 3x^3/x + c
= x^2 – x^3 + c this answer is called an anti-derivative
2. S1/2x^3dx = S1/2x^-3dx
= 1/2x^-2/-2 + c
= -1/4x^2 + c
3. S(t^2-cost)dt
= 1/3t^3 + sint + c

Helpful-Derivative of: sectan > sec
sec^2 > tan
csccot > csc

bSa is where it starts and stops with area.
To find the fundamental theorem of calculus:
F(b) – F(a)
a> smaller #
b> bigger #
Examples:
1. Find the area of the region bounded by the graph of y=2x^2-3x+2, the x-axis and the
vertical x=0 and x=2.
2S0 2x^2-3x+2
2[x^3/3] – 3[x^2/2] + 2x
(2/3(2)^3 – 3/2(2)^2 + 2(2)) – 2/3(0)^3 – 3/2(0)^2 + 2(0)
= 10/3
Then you would graph the equation to see what is between 2 and 0.

2. y=cosx
pi/2S0
sinx
sin(pi/2) – sin(0)
= 1

Week 3 Blog Prompt

Create and workout your own example problem for optimization. Be sure to explain each step that you take in the solution and why.

blog on MLK day

we have a test this week on this stuff, so a review is much needed

This is how you solve differentiable. Differentiate both sides with respect to x. d_/dx
2) Collect all dy/dx terms on one side, and move the other terms to the other side.
3) Factor out dy/dx.
4) Solve for dy/dx.
5) Simplify

To do the first derivative you do the following:
1.take the derivative. set it equal to zero.

2.find anywhere that it is not differentiable.

3. set up intervals using the first two steps of the
process.

4. plug in a number from each interval you
found into the derivative equation.

5. if it is +ve it is increasing.
if it is -ve it is decreasing.

6. determine the max or min from
step 5
increasing - max
decreasing – min
I found otimazation was really easy, but these are the steps

1. identify all given quanities and what
you are to find.

2. find a primary equation:
**hint-it is the one that is closest
to the word maximize or minimize.

3. find a secondary equation if necessary
(only if there are two variables in the
equation in step two)

4. find a max or min by plugging into
the primary equation and do the first
derivative test. instead of intervals
plug back into primary equation to
find the highest and lowest value

Lindsey's Reflection

This week we learned about optimization.

here are the steps in solving an optimization
problem:

1. identify all given quanities and what
you are to find.

2. find a primary equation:
**hint-it is the one that is closest
to the word maximize or minimize.

3. find a secondary equation if necessary
(only if there are two variables in the
equation in step two)

4. find a max or min by plugging into
the primary equation and do the first
derivative test. instead of intervals
plug back into primary equation to
find the highest and lowest value.

here is an example applying these steps:

Find two non-negative numbers whose
sum is 9 and so the product of one number
and the square of the other number is
a maximum.

1. a + b=9 P=a x b^2

2. p= a x b^2--primary equation

3. a + b=9 b=9-a

4. p= a(9-a)^2

p= a (81-18a+a^2)
=81a-18a^2+a^3

81+36a+3a^2=0
3(a^2-12a+27)=0
(a-3) (a-9)

a=3,9

3=108--max
9=0

108=3 x b^2

a=3
b=6

not too hard. goodnight.

Lindsey's Reflection

MLK Jr Reflection

Ok its time to roll out for the night so i'ma come hit up this blog right quick on some old stuff cuz no time to figure out the new stuff an put it on here. Implicit derivative, here we go:
Steps:
1) Differentiate both sides with respect to x. d_/dx
2) Collect all dy/dx terms on one side, and move the other terms to the other side.
3) Factor out dy/dx.
4) Solve for dy/dx.
5) Simplify.

And an example of course:
x^2 + y^2 = 0
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
= -x/y


Alright good people I'm out...everybody enjoy tonight an Monday off tomorrow!!!

Terrio's Reflection

So I'm going eat at LaCarreta's in like 20 minutes and I need to hurry up and get this over with before I forget....kind of like I forgot my math stuff once again. I'll just review some old stuff that I remember and still have scratch work from on my computer desk from the past....

Implicit and explicit derivatives :
explicit-solved for y
implicit-not solved for y
When you take the derivative, you must put the notation, dy/dx, by any variable
except for x.

Example :

1) x^2 + y^2 = 9
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
= -x/y

When solving related rates problems, you must follow the guidelines.

Example :
1) y = square root(x)
dy/dt = ? x = 4 dx/dt = 3
y = square root(x)
dy/dt = 1/2x^-1/2
dy/dt = 1/2(4)^-1/2
= 3/4

Reflection

This week in calculus, we learned how to solve optimization problems.
STEPS:
1. Identify all given quantities and what you have to find.
2. Find a primary equation.
3. Find a secondary equation if necessary. (Only if there are 2 variables in
equation in step 2.)
4. Find max or min by plugging into primary equation doing first derivative test.
Instead of intervals plug back into primary equation to find highest and lowest
value.

EXAMPLE from notes:
Find two numbers whose sum is 10 and product is as large as possible.
1. a+b=10
2. P=ab > primary
3. a+b=10
b=10-a
4. P=a(10-a)
=10a-a^2
Derivative: 10-2a=0
a=5
5 > 10(5)-5^2=25
P=ab
25=5(b)
b=5

EXAMPLE from quiz:
A farmer has 2400 ft of fencing. What are the dimensions of a large area of the pen.
1. P=2400
l>b
w>a
2. A=ab
3. 2a+2b=2400
4. 2a=2400-2b
a=1200-b
A=(1200-b)b
=1200-b^2
1200-2b=0
-2b=-1200
b=600
A=1200(600)-600^2
A=3600
A=ab
3600=600a
a=6000
> 600 X 600

First reflection on the 2nd semester

So this week we took it pretty easy we just reviewed some stuff. One of the things we reviewed was the first derivative test……an that’s what this blog’s on.
STEPS:

1. take the derivative. set it equal to zero.

2. find anywhere that it is not differentiable.

3. set up intervals using the first two steps of the
process.

4. plug in a number from each interval you
found into the derivative equation.

5. if it is +ve it is increasing.
if it is -ve it is decreasing.

6. determine the max or min from
step 5
increasing - max
decreasing - min

here an example:

find the relative extrema of
f(x) = (x^2 - 4)^2/3

1. 2/3 (x^2-4)^-1/3 (2x) = 0

4x/3 (x^2-4)^1/3 = 0

4x = 0

x = 0

2. x^2 – 4 = 0
critical point = -2, 0, 2

3. (-infinity,-2)u(-2,0)u(0,2)u(2,infinity)

4. f^1(-3) = -ve dec
f^1(-1) = +ve inc
f^1(1) = -ve dec
f^1(3) = +ve inc

max: x = 0

min: x = -2, 2

Terrio's Reflection

This past week was a review for the most part. I think we start new stuff either on Monday or Tuesday depending on whether or not we go to class on Monday. This is what we did....

Rolles Theorem :
1. Check to see if it is continuous .
2. If so check if it is differentiable .
3. Make sure the y-values match .
4. Take the derivative and set it equal to 0, and solve for x.

Ex :

2 x ^ 2 + 4 x ( - 1, -1 )

1 . yes
2 . yes
3 . f( - 1 ) = 2 (-1) ^ 2 +4 (-1) = - 2
f( - 1 ) = 2 (-1) ^ 2 +4 (-1) = - 2
4 . 4 x +4 = 0
4x = - 4
x = - 1

First Derivative Test :
1 . Take derivative and set it equal to zero.
2 . Find anywhere it is not differentiable
3 . Set up intervals using steps 1 and 2.
4 . Plug in a number on the interval into the derivative equation
5 . If it is positive it is increasing, if negative it's decreasing
6 . determine max or min.
If increasing it is a max
If decreasing it is a min

Lindsey's Reflection

This week we just reviewed
the first derivative test and the
second derivative test.

here is the steps to the first derivative
test:

1. take derivative and set it equal to
zero.

2. find anywhere it is not differentiable

3. set up intervals using the first two steps.

4. plug in a number on the interval into
the derivative equation.

5. if it is positive it is increasing if
it is negative it is decreasing

6. determine which is the max and which is
the min using step five

increasing- max
decline-min

here are the steps for the second derivative
test:

1. take the second derivative and set it
equal to zero

2. find anywhere it is not differentiable

3. set up intervals using steps one and
two

4. plug in a number on the interval into
the 2nd derivative equation

5. if it is positive it is concave up and if
it is negative it is concave down

6. if changes from + to - or - to +
there is a point of inflection.

We also learned the shortcut to the
first derivative test this week:

1. take the derivative and set it equal
to zero. solve for x

2. take the second derivative.

3. plug in x values from step 1.
-if positive- min
-if negative-max

**the test fails if there is no second
derivative

here is an example using the shortcut:

3x^5+5x^3

1. -15x^4+15x^2=0
15x^2(-x^2+1)=0

x=0, +/- 1

2. -60x^3+30

f(0)=0 --the test fails
f(1)=- max
f(-1)=+-min

Reflection

This week we reviewed from a previous chapter, these are the steps to rolle’s theorem, finding extrema, max or mins.

Rolle Theorem:
1. Make sure it is continuous.
2. Make sure it is differentiable.
3. Make sure the y-values match.
4. Take the derivative, set = 0, and solve for x.
example of the Rolle Theorem.
EXAMPLE:
Let f(x) = x^4 - 2x^2
Find all values c in the interval (-2, 2) such that f'(c) = 0.
1. ok
2. ok
3. f(-2) = (-2)^4 - 2(-2) = 8
f(2) = (2)^4 -2(2)^2 = 8
4. f'(x) = 4x^3 - 4x = 0
4x(x^2 - 1) = 0
x = 0, + or – 1

Steps for finding extremas, maxs, or mins:
1. take the derivative and set it equal to zero. Solve for x.
2. Find anywhere the function isn’t differentiable.
3. plug in to get a y-value if the problem is asking
about a max or min specifically.
4. Set up intervals
5. the highest number is the max.
the lowest number is the min.

First Derivative Test:
1. take derivative and set it equal to zero.
2. find anywhere it is not differentiable
3. set up intervals using steps 1 and 2.
4. plug in a number on the interval into the derivative equation
5. if +ve it is increasing. if it is -ve it is decreasing
6. determine max or mins.
increasing- max
decreasing- min

holiday blog 3

in this blog i will go over stuff we did right before the holiday, well kinda before the holidays, im more refreshing the first derivative test cause i know we will be asked this and i want to BAM knock her dead with my answer!
1. take the derivative. set it equal to zero.
2. find anywhere that it is not differentiable.
3. set up intervals using the first two steps of the
process.
4. plug in a number from each interval you
found into the derivative equation.
5. if it is +ve it is increasing.
if it is -ve it is decreasing.
6. determine the max or min from
step 5
increasing- max
decreasing-min
here an example:
find the relative extrema of
f(x)=(x^2-4)^2/3
1. 2/3(x^2-4)^-1/3(2x)=0
x/3(x^2-4)^1/3=0
4x=0
x=0
2. x^2-4=0
critical point=-2,0,2
3. (-infinity,-2)u(-2,0)u(0,2)u(2,infinity)
4. f^1(-3)=-ve-dec
f^1(-1)=+ve-inc
f^1(1)=-ve-dec
f^1(3)=+ve-inc
max: x=0
min: x=-2,2

holiday blog 2

The next review is probably the most important one of all. We cannot forget how to do the product and quotient rule!

Both of these rules include a formula.
here is the formula for the product rule:

(recopy the first equation)(take the derivative of the second equation)+(recopy the second equation)(take the derivative of the first equation)

here is a example where you would have to
apply this rule:

(3x-2x^2)(5+4x)

since these two equations are being
multiplied you have to use the product
rule and plug in the formula

After doing so this is what you end up with

(3x-2x^2)(4)+(5+4x)(3-4x)

then you simplify as much as
you can

12x-8x^2+15-20x+12x-16x^2

=-32x^2+4x+15


next is the quotient rule
here is the formula:

(bottom)(derivative of top)-[(top)(derivative of bottom)]/ (bottom)^2

the only way that this rule would not apply is if
you have a number instead of an equation
at the bottom of the fraction.

anyway here is an example using
the quotient rule:

5x^2-2/x^2+1

(x^2+1)(5)-[(5x-2)(2x)/(x^2+1)^2

from here, once again, you just
simplify as much as possible

5x^2+5-[10x^2-4x]/(x^2+1)^2

always keep the bottom of the problem
as is.

5x^2+5-10x^2+4/(x^2+1)^2

=-5x^2+4x+5/(x^2+1)

holiday blog 1

All the following blogs will be a review so that when i got back to calc again i wont be total lost!
Starting with the beginning of the year
lim 2x+4
x-->2

for this problem it is not division or anything to where you would get in the bottom of the problem.

there fore you can easily plug in for this equation and get an acceptable answer.

lim 2x+4
x-->2


you would plug the 2 in where the x is to get the answer.

2(2)+4

4+4=8

the limit for this problem would be 8..

this for me is the easiest type of limit problem to solve but
unfortunately each time will not be as easy as the next.

in some problem you have to plug the problem into a table and find your answer that way.


this is the graph.


x -.1 -.01 -.001 .001 .01 .1

this is what you plug the problem into.

you have to add the number given to all of the numbers.

after doing so you plug the whole equation in to your graphing calculator and the table that you have set up will give you the answers for each different number that you have figured in the table.

most of the time both sides will be approaching a specific number and that will end up being the limit of the problem. you have to compare both sides of the problem and this will only work if both sides that you have compared are the same or they match each other.

in the case that the right and the left side are not the same or do not match..
then the limit does not exist and this will be the answer of the problem!

Holiday Reflection #3

Here are some formulas and theorems to remember for after the holidays. They mostly deal with taking derivatives.


Steps for finding extremas, maxs, or mins:

1. Take the derivative and set it equal to zero. Solve for x.
2. Find anywhere the function isn’t differentiable.
3. Plug in to get a y-value if the problem is asking about a max or min specifically.
4. Set up intervals.
5. The highest number is the max & the lowest number is the min.

*When asked to find critical numbers, you just do the first two steps.

Rolle's Theorem:

1. Make sure the function is continuous.
2. Make sure the function is differentiable.
3. Make sure the y values match.
4. Take the derivative and set it equal to zero, then solve for x.

Mean Value Theorem (MVT):

-Similar to Rolle’s Theorem but after you do the first to steps you have to set the derivative equal to the slope of the function and solve for x.

Holiday Reflection #2

L I M I T S !

-They NEVER go awayyyyy!! So i'm going to review limits to make sure i'm ready for when we get back.

lim
x->infinity

1. Degree at top = degree bottom lim is the coefficient.
2. Degree top > degree bottom lim is infinity or -infinity.
3. Degree top < degree bottom lim is 0.

EXAMPLE:

lim (2x+5)/(3x^2+1)= 0
x-->infinity

*The answer is 0 because the top degree is less than the bottom degree.
*And if worse comes to worse, you can always use the table, but make sure to see if you can't do it a shorter way because the table takes longer.

Holiday Reflection #1

Quotient Rule and Product Rule?? I believe so. These are the lessons that I remember most from the second nine weeks. So let's review them.

Product Rule: (copy 1st)(derivative of 2nd)+(copy 2nd)(derivative of 1st)

Quotient Rule: (copy bottom)(derivative top)–[(copy top)(derivative bottom)]/( bottom ) ^2

PRODUCT RULE:

(2x^2+5) (3x^2-1x)
= (2x^2+5)(6x-1)-(3x^2-1x)(4x)
= 12x^2-2x^2+30x-5 - 12x^2-4x^2
= -6x^2+30x-5

QUOTIENT RULE:

x^2 + 3 / 2x
= (x^2 + 3)(2)–(2x)(2x)/(2x)^2
= 2x^2–4x+3/(2x)^2

Well that was easy! I hope that these are on EVERY test, because these problems are easy points. Just l e a r n the formulas! (:

Holiday Reflection #3

LAST HOLIDAY BLOG!!!

So it will be yet another review blog. This is a review on Rolle’s Theorem…or is it Roole’s Theorem? Well that is unimportant, here’s how ya do it:
Use these steps:

1.) Make sure the function is continuous
2.) If it is continuous make sure that it is differentiable
3.) Make sure that the y - values match
4.) Take derivative and set equal to 0, then solve for x

Here is an example:

x^2 - 2 x Find all values c in the interval ( -2, 2 ) such that f'( c ) = 0
1. yes
2. yes
3. f( -2 ) = -2^2 - 2( -2 ) = 8
f( 2 ) = ( 2 )^2 - 2( 2 ) = 8
4. 2x – 2 = 0
2x = 2
X = 1
So x = 0 is your min and x = 8 is your max because they are the greatest an least out of all your values

Holiday Reflection #2

For my second holiday blog, i will do somethin a LITTLE more recent than product and quotient rules...Here is a little review on some stuff from the beginning of chapter 5, NATURAL LOGS!

First off, here is how you expand and condense natural logs:

Expand:
ln xy^2/5
lnx + lny^2 – ln5
= lnx + 2lny – ln 5

Condense:
2ln(x+3) – ln(x-4)
= ln (x+3)^2/(x-4)


Some notes to remember when fooling with logarithms:
To solve for a variable when it’s an exponent:
1. Take the ln of both sides.
2. Bring the variable to the front of the ln.

To solve for a variable if it’s inside a log:
1. Rewrite as an exponent.
2. Solve for x.

And here are some examples of this:

6 = e^x-2
ln6 = lne^x-2
ln6 = x-2
x = ln6-2

ln(3x-2) = 9
e^9 = 3x-2
e^9+2 = 3x
x = e^9+2/3

e^x = 12
lne^x = ln12
x = ln12
= 2.485

Holiday Reflection #1

Well for my first holiday blog, why not start out with somethin extremely easy from way back when? Yep thats what i'm gunna do....I'ma do a reflection on the Product Rule and the Quotient Rule. These are the formulas in simple terms for these two rules:
Product Rule: ( copy 1st ) ( derivative of 2nd ) + ( copy 2nd ) ( derivative of 1st )
Quotient Rule: ( copy bottom ) ( derivative top ) – [( copy top ) ( derivative bottom )] / ( bottom ) ^2
It is much simpler if you look and the formula and an example at the same time, so check this out!
Find the derivative of (3x^2+4) (2x^2-3x)
= ( 3x^2+4 ) ( 4x-3 ) - ( 2x^2-3x ) ( 6x )
= 12x^3 - 9x^2 + 16x - 12 - 12x^3 + 9x^2
= 16x – 2
And one for the quotient rule of course:
X^2 + 3 / 2x
= ( x^2 + 3 ) ( 2 ) – ( 2x ) ( 2x ) / (2x)^2
= 2x^2 – 4x + 3 / (2x)^2