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