Reflection #14

we were on a holiday so im going to review stuff:
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)

so it is not too bad as long as you
memorize the formula!

Week 6 Prompt

Give an example of each type of differentiation we did in Ch. 5. Use each formula and make sure your example problem illustrates how the formula is used.

Reflection #14

For this blog I will do a review of something boo koo easy and pretty old, but oh well I’m doin it any way. The product rule and the quotient rule are two formulas that refuse to go away in calculus, so if you not good at them yet, study this blog post!
Product Rule: ( copy 1st ) ( derivative of 2nd ) + ( copy 2nd ) ( derivative of 1st )
Quotient Rule: ( copy bottom ) ( derivative top ) – [( copy top ) ( derivative bottom )] / ( bottom ) ^2
-Hear is an example of the product rule:
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
-and the quotient rule:
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

So i forgot to do my first blog but here's my second.

since we were off we didnt do anything new
so this blog is going to be about the first derivative test.

here are the steps to this process:

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

thats all to it. see everyone tomorrow.

Reflection #14

The Holidays go by way too fast. Here's a review on how to do the shortcut to taking the derivative of simple equations.

- Rule number 1 -Constant Rule - the derivative of any constant is always zero. Rule number 2 - Power Rule - take the exponent and bring it to the front of the variable, then take the exponent and subtract it by one.

- Example of Constant Rule: y = 7 the constant rule says the derivative of any constant is zero, so the derivative is zero.

- Example of Power Rule: f ( x ) = x^6 move the 6 to the front and subtract one from the exponent and your derivative would be 6 x^5

- SHORTCUT FOR SIN AND COS:

If you have cos(x) then the derivative is sin( x ) . When you have sin( x ) the derivative is -cos ( x )

- Example :

y = 8 cos ( x ) You change the cos( x ) to - sin ( x ) and you would get y = 8 ( - sin ) ( x ) . Then you get y = -8 sin ( x ) . So the derivative is y = -8 sin ( x ) .

Reflection

This week was our holiday, so I'll do a blog to review some of the stuff we learned throughout the year. We have learned information from chapters 1, 2, 3, and 5, but out of all of them, I think 3 was the most difficult.

2.5 Implicit Derivatives
When solving for implicit derivatives, it is easiest to solve by following 5 steps.
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.
EXAMPLE:
x^2 + y^2 = 0
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
= -x/y

Chapter 1
EXAMPLES:
1) lim tanx/x = sinx/cosx/x/1
x>0
= (sinx/x)(1/cosx)
= (1)(1/cos(0))
= (1)(1)
= 1

2) h(x) = -x^2 + 4x
a) lim h(x) = -(4)^2 + 4(4) = 32
x>4
b) lim h(x) = -(-1)^2 + 4(-1) = 5
x>-1

Week 6 Prompt

Due to the fact that the internet was not dependable on my laptop I was unable to post the blog prompt for the holidays. Therefore it is a freebie for everyone. Just make sure to post your 2 regular blogs. They can come from any review topic.

Reflection #13

So glad the thanksgiving hoidays are here but i will be busy practicing for curtis, i hope we can pull off a win and happy thanksgiving to all.
last week we learned this.....
- How to Expand
- How to Condense
- How to take a derivative of a natural log
EXAMPLES
Expand.
1) ln xy^2/5
lnx + lny^2 – ln5
= lnx + 2lny – ln 5
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.

EXAMPLES
1) 6 = e^x-2
ln6 = lne^x-2
ln6 = x-2
x = ln6-2

Reflection #13

This week was LOOOOOGGGGGG week. we learned a buncha stuff about logs this week, well, mostly natural logs, but it wasnt too bad....

how to expand natural logs:

1) ln xy^3/10
ln x + ln y^3 – ln 10
= ln x + 3ln y – ln 10

how to condense logs:

3ln x
= ln x^3

how to take the derivative of a natural log:

-Use this formula: d / dx ln u = 1 / u * d (u)

ex: ln x^2 - 1
1 / x^2 - 1 * (2x)
2x / x^2 - 1

BTW I did these blogs at a party hahaha

Terrio's Reflection

This week we learned this.....

- How to Expand
- How to Condense
- How to take a derivative of a natural log

Expand :

lnx^2y/10
=2 ln x + ln y - ln 10

Condense :

3 ln ( x + 2 ) - ln ( x - 1 )
= ln ( x + 2 ) ^3 / ( x - 1 )

Taking a Derivative :

d / dx ln u = 1 / u * d (u)

d / dx ln 2 x
=1 / 2x * 2
=1 / x

Reflection

This week in calculus, we worked with solving natural logarithms by using different methods like solving for derivatives.
Section 5.1
EXAMPLES
Expand.
1) ln xy^2/5
lnx + lny^2 – ln5
= lnx + 2lny – ln 5

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

Use ln3 ~ .49 and ln4 ~ .23 to approximate ln12
3) ln4/3 = ln4 – ln3
= .23 - .49
= -.26

Section 5.4
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.

EXAMPLES
1) 6 = e^x-2
ln6 = lne^x-2
ln6 = x-2
x = ln6-2

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

3) e^x = 12
lne^x = ln12
x = ln12
~ 2.485

Week 5 Blog Prompt

Using Calculus how can you find a max or a min? How do you determine if the value is a max or min?

I think it's E=MC squared...........? hahahahahaha

YEAH YEAH YEAHHHHHHHHHHH

CALCULUSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS!!!!!!!!!!!!!!! Yeahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, this is what we do when we don't feel like actually workingggggggggg!

E=MC...you know

ummmmmmm, nope. I dont think I did

We want to make blogging FUN:)

I LOVE MY MATH, I REALLY LOVE MY MATH, MATH IS MY LIFE!!!!!!!!!! I'm high on math. WOAH!!!!!!!!!!!!!!!!!!!!

you created the puddle...with your peeeeeeeeee

hahaha yeah definitely, idk what that puddle was though....

???????

do you think i should of dropped sarah in that nasty puddle in the road haha

dude idk know but i do know that we were jammin and that coach thought we were on crack

Mannnnnnnnn, I'm gunna be sore tomorrow. Manday killed me. Why did we wake up at 5 in the morning to go workout for 4 hours?

HEY!!!!

MANDAY WAS INTENSE TODAY!! WE DONT EVEN WAKE UP THAT EARLY FOR SCHHOL/CHURCH

This weeks notes:

- 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.
*You use Rolle Theorem when trying to find a max or a min, or trying to prove that there is a max or a min.

- Mean Value Theorem:
*Set derivative = to slope between two points and there has to be a value x you can solve for if it is continuous and differentiable on the interval.

- First Derivative Test:
1. Take derivative = 0.
2. Find where it is not differentiable.
3. Set up intervals using step 1 and 2.
4. Plug in a number on the interval into the original equation.
5. If positive, it is increasing, if negative, it is decreasing.
6. Determine max or mins from step 5. Increasing / decreasing - max; Decreasing / increasing - min.

- Ex of a problem you may get :

Find the absolute extrema of ( 4 / 3 x + 5 ) at ( 0 , 5 )

Do all the steps. Take derivative and set equal to zero and you would get ( 12 / 16 ) = 0
That derivative does you no good, so you plug in 0 and 5 to the original equation separately. When you plug in 0 you get ( 4 / 5 ) and when you plug in 5 you get 5.
So your answer is Max : ( 5 , 5 ) Min : ( 4 / 5 )

Reflection #12

I suck at this stuff and I’ma prolly bomb the test tomorrow, but here’s some stuff you can study:

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 trying to find critical numbers you only do the first 2 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:

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.

Reflection

This week I’m going to review how to solve for a derivative using the long way. Two things you need to know if you’re trying to solve for a derivative and your looking at a graph is a secant line and a tangent line. A secant line is a line that touches a graph in one spot. A tangent line touches the graph in two spots. When trying to solve for a derivative you use the formula f(x+∆x)-f(x)/∆x to help you solve it. When the directions say y', f'(x), dy/dx, d/dx, Dx[y], slope of a tangent line, or slope of the curve at a point it is telling you to find the derivative. Let’s try an example:
Find dy/dx of x²+1 at the points (0,1) & (-1,2)
=(x+∆x)²+1-(x²+1)/∆x
=x²+2x∆x+∆x²+1-x²-1/∆x
=2x∆x+∆x²/∆x
=∆x(2x+∆x)/∆x
=2x+∆x
=2x
Then you take the two points and plug it into the equation
Slope at (0, 1) = 2(0) = 0
Slope at (-1, 2) = 2(-1) = -2

Lindsey's Reflection

So we have a test tomorrow.
im super nervous but this is what
its on!

Steps for finding extremas, maxs, or mins:

1. take a derivative and set it equal to zero.

2. find any place the function is not
differentiable.

3. plug in to get y-value if asking
about a max or min specifically.

4. if on an interval plug in to get
a y value.

5. the highest number is the max.
the lowest number is the min.

if you are looking for just extrema then only
use the first two steps but if you are looking
for max and min you have to use every
step.

Roolle's Theorem:

1. make sure it is continuous.

2. make sure it is differentiable.

3. make sure that the y values match.

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

you use roolle's theorem when trying to find
a max or min or when you are trying to prove
there is a max or a min.

Mean Value Theorem:

-set derivative equal to slope
between 2 points and there has to be
a value x. you can solve for it if it is continuous and
differentiable on the interval.

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

Reflection #12

Well we have a test tomorrow and me matt and terrio are having a hardcore review right now.
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.
*You use Rolle Theorem when trying to find a max or a min, or trying to prove that there is a max or a min.
Mean Value Theorem:
*Set derivative = to slope between two points and there has to be a value x you can solve for if it is continuous and differentiable on the interval.
First Derivative Test:
1. Take derivative = 0.
2. Find where it is not differentiable.
3. Set up intervals using step 1 and 2.
4. Plug in a number on the interval into the original equation.
5. If positive, it is increasing, if negative, it is decreasing.
6. Determine max or mins from step 5. Increasing/decreasing - max; Decreasing/increasing - min.
*critical points= step 1&2
*extrema,max,min,inc.,dec.,= all 6 steps
Steps to Related Rates Problems.
-Identify all given quantities and quantities
to be determined. Make a sketch and label the
quantities, or Identify what you are given in the problem.
-Write an equation with the variables
whose rates of change are either given or are
to be determined.
-Using the Chain Rule, implicity differentiate
both sides of the equation with respect to
time t. Meaning rates will be (d__/dt)
-Last, substitute into the equation all the known values for the variables and their rates of change. Then solve for the rate of change you are asked for.
Ex:xy^3=3
dy/dt=?x=5 dx/dt=4
xy^3=3
3xy(dy/dt)+y^2(dx/dt)=0
3(50y(dy/dt)=y^3(4)
15y(dy/dt)=3y^2
dy/dt=4y^2/15y

Reflection

For my blog this week, i'm going to review chain rule, veritcal asymptotes, and derivatives. To begin, 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

Second, when solving chain rule, you take the derivative from outside in. An easy way to remember how to do this is by saying the formula like this: derivative of outside, recopy inside, multipy 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

Finally, when solving for derivatives, it is usually a simple process. Derivatives can be solved different ways, 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

Week 4 Blog Prompt

How can you find the domain of a graph? How do you know if a functions is increasing or decreasing? Give an example of each.

Prompt?

Well it's Wednesday and B-Rob hasn't been to school for a while anddddd we don't have a prompt up. So, Math is cool haha peace.

Reflection 11/7

alright, so i'm not too sure if this is supposed to be easy or not, but it definitely is not easy for me. I missed one day of school, i'm not sure which day, but i missed the Rolle Theorem lesson. So i have absolutely no idea how to do that section of the chapter. If anyone can comment to help explain, then i might not fail the test when we have it. But i can try to explain the notes and maybe it will help me learn.

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.

*You use Rolle Theorem when trying to find a max or a min, or trying to prove that there is a max or a min.

Mean Value Theorem:

*Set derivative = to slope between two points and there has to be a value x you can solve for if it is continuous and differentiable on the interval.

First Derivative Test:

1. Take derivative = 0.
2. Find where it is not differentiable.
3. Set up intervals using step 1 and 2.
4. Plug in a number on the interval into the original equation.
5. If positive, it is increasing, if negative, it is decreasing.
6. Determine max or mins from step 5. Increasing/decreasing - max; Decreasing/increasing - min.

*critical points= step 1&2
*extrema,max,min,inc.,dec.,= all 6 steps


Well that is all we learned for the entire week, and i think i understand most of it. But more practice will surely help. It is much easier when we are working problems in class.

Reflection #11

Well matt couldnt of said it better. We at least tried to do as much as we could but i found this week to be very difficult. This week we learned a lot about Rolle's Theorem but I am not too good at it but I kind of get the general concept. Rolle’s Theorem is used when you are trying to find a max or a min.

The steps are
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

the highest is the max and the lowest is the min.if you are looking to find an extrema he first two steps are all you need to use do is fina and max and a min you have to use all five steps.

we also learned how to use the mean value theorem. You use this when the average “rate of change” the rules that you need to use when you are using this theorem:set the derivative equal to zero to the slope between two points and there has to be a value x you can solve for if it is continuous and differentiable on the interval.

Reflection #11

This week was a beast, an most of what I know came from doin calc wit terrio an Chauvin so I’m gunna do a blog really similar to Terrio, and Chauvin will prolly do one similar to me an Terrio’s. This week we learned a lot about Rolle's Theorem but I am not too good at it but I kind of get the general concept. Rolle’s Theorem is used when you are trying to find a max or a min.

These are the 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

Ex :

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? this is the part I’m shaky on.

Lindsey's Reflection

this week we started chapter 3.
the first thing we learned about was
finding extremas and max and mins.

there are steps to doing so.
here they are:

step1:

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

step2:

find any place the function
is not differentiable.

step3:

plug in to get a y value
if asking about a max
or a min specifically.

step4:

if on an interval plug
in to get a y value for
each.

step5:

the highest is the max
the lowest is the min.

if you are looking to find an extrema
the first two steps are all you need to
use. but in order to find and max and a
min you have to use all five steps.

here is an example:

find the extrema of
f(x)= 3x^4-4x^3 on [-1,2]

1. 12x^3-12x^2
= 12x^2(x-1)=0
x=0,1

2. ok

3. f(0)=0
f(1)=-1-min

4. f(-1)= 7
f(2)=16-max


thats all there is to it and if you
were just looking for the extrema
like i said all you would look
at is the first two steps and in
this case the extrema or critical numbers
would be 0 and 1.

goodnight.

Reflection #11 Terrio

This week we learned chapter three sections one and two. We learned Rolle's Theorem and how to do this I'm not really sure because I haven't been asking questions when I get lost this week...I was lost the whole week so I'll try to explain this as best I can. You use Rolle's Theorem to prove whether or not there is a max and a min.

Steps:

1.) Mak sure it is continuous
2.) If so make sure it is differentiable
3.) Make sure y - values match
4.) Take derivative equal; solve for x

Ex :

x^4 - 2 x^2 Find all values c in the interval ( -2, 2 ) such that f'( c ) = 0
1. yes
2. yes
3.f( -2 ) = -2^4-2( -2 )^2 = 8
f( 2 ) = ( 2 )^4-2( 2 )^2 = 8
4. 4 x^3-4 x = 0
4 x ( x^2-1 )
x = 0, -1, 1

Reflection

This week we learned section two and three in chapter three. We learned how to use Roole’s Theorem and when you can use it. You can only use Roole’s Theorem when you are tring to find a max or min or trying to prove there is a max or a min. the rules are:

1. make sure its continuous
2. make sure its differentiable
3. make sure the y values match
4. take the derivative and set it equal to zero and solve for x.

we also learned how to use the mean value theorem. You use this when the average “rate of change” the rules that you need to use when you are using this theorem:

set the derivative equal to zero to the slope between two points and there has to be a value x you can solve for if it is continuous and differentiable on the interval.

Reflection

This week in calculus we began learning Chaprer 3. In Section 3.1 we learned how to find a max or min, and its is the same as finding a horizontal tangent line. When working a problem like this there are steps that must be followed.
STEPS:
1) Take derivative and set equal to 0.
2) Find any place that the function is not differentiable.
3) Plug in to get y-value if asking about max or min.
4) If on an interval plug in to get a y-value.
5) Highest-max, lowest-min

In Section 3.2 we learned about the Rolle Theorem and the MVT Theorem, but I'll give an 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

Week 3 Prompt

What are the rules for simplifying exponents? What are the rules for simplifying logs?