Reflection #10

This week was alright i guess, we had a sub on thursday and friday, but i was out thursday and friday i worked on the take home test which by the way is extremely difficult!!! I did alot of related rate problems so ill talk about this.

1. Identify all given quantities and quantities
to be determined. Make a sketch of the problem and label the
quantities correctly.

2. Write an equation involving the variables
whos rates of change either are given or are
to be determined. In other words, list what you KNOW, and list what you are TRYING TO FIND.
3. Using the Chain Rule, implicity differentiate
both sides of the equation with respect to
time t. (this means that all rates will be d something/dt
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.

Key words to look for are: Cubic = looking for volume; the rate of = d something/ dt

Example:
xy = 6
Step 1: dy/dt = ? x = 4 dx/dt = 8
Step 2: xy = 6
Step 3: xdy/dt + ydx/dt = 0
Step 4: xdy/dt = -10y
= -2y

Reflection #10

I was gone most of last week so I don't think we really learned anything new so I'll talk about something I hate even tho i do understand it pretty well. This thing of which I speak of in calculus terms is called finding the " implicit derivatives. " This is a pretty simple concept to understand. The only difference between implicit and a regular derivative is that in an implicit derivative, every time you take the derivative of something with a " y " in it you have to put a " dy / dx " behind it. After you do that you then solve for dy / dx and wah lah, you have taken an implicit derivative.

Ex. Find the implicit derivative of 2x^2 + 2y = 6

-take the derivative of the function and add the dy / dx:

4x + 2 dy/dx = 0

-solve for dy / dx:

2 dy/dx = -4x
dy/dx = -2x

See, it is that simple!!!!

Terrio's 10th Reflection

This week was mostly a big review for the Test we took on Wednesday and then Thursday and Friday we had Mrs. Fe Fe as a sub and we worked on the chapter test review and the take home test that we got on Monday. One thing I noticed we did a lot this week, however, was solving Implicit Derivatives. To solve Implicit Derivatives you basically take the Derivative and solve for dy/dx.

Steps to solve an Implicit Derivative are....
- Differentiate both sides (terms where the derivative of y are taken need a dy/dx behind them)
- Get all the dy/dx terms to one side and the other terms to the other.
- Factor out dy/dx
- Solve for dy/dx
- Simplify your answer

Ex. x^2 + y^2 = 239823502194568209345620345862093457120946510

- 2x + 2y dy/dx = 0
- 2y dy/dx = -2x
- dy/dx ( 2y ) = -2x
- dy/dx = -2x / 2y
- dy/dx = -x / y

This week was mostly a review on how to solve problems when we have implicit derivatives, explicit derivatives and a second derivative in an implicit function. To solve for an implicit derivative and a explicit derivative you have to know the difference. An implicit derivative is not solved for y and an explicit derivative is solved for y. when your solving for either the implicit or explicit derivatives after you take the derivative of something with a y you have to put the dy/dx notation behind it.

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

Lindsey's Reflection

We recently had a test on chapter 2.
this test covered such things as derivatives, differentiation, product rule and quotient rule, the chain rule, and implicit differentiation.


here are a few notes on these concepts:

Derivatives:

there is a long way and a shortcut.

all of these mean you are dealing with a derivative:
y^1, f^1(x), dy/dx, d/dx, slope of a tan line,
slope of the curve at a point.

Continuous:

a function is continuous if it doesnt contain
one of the four types of discontinuities.

Types of discontinuities:

1. removable
2. jump
3. infinate
4. oscillation


Chain rule:

-take the derivative from the outside in

-derivative of outside, recopy inside, multiply by derivative of inside.

Implicit:

implicit- not solved for y

explicit-solved for y

Reflection

In calculus this week before we took another chapter test we learned more about word problems, related rates, and implicit and explicit derivatives.

2.5 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.
+ Examples:
1) x^2 + y^2 = 9
2x + 2ydy/dx = 0
2ydy/dx = -2x
dy/dx = -2x/2y
= -x/y
2) xy = 7
(x)(dy/dx) + (y)(1) = 0
xdy/dx + y = 0
dy/dx = -y/x

2.6 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

Word Problems
+ Example:
1) 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.

Week 2 Blog Prompt

What are the different ways a derivative can be interpreted as? So far what have you used the derivative for?

Reflection #9

This week was kind of a tough one for me, i still have to go and try to finsh those 100 homework problems. We did some stuff on related rates.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 #9

This week we started reviewing chapter 2 for our test on Wednesday. We learned word problems too, but I’m not really comfortable enough with that to do a one hundred fifty word blog on it. So instead I’ll review a few of the things that we learned the past few weeks in chapter 2.
Find the equation of a tangent line:
-when doing this you have to take the derivative of an equation, then plug in your values then plug the derivative into point slope form.
Ex. 2x^2 + 4x + 2 @ ( 2, 3 )
First take the derivative of the equation: 4x + 4
Now plug in your x value: 4 ( 2 ) + 4 = 12
Now plug into point slope: y – 3 = 12 ( x – 2)
SIMPLE!
Now a review of the chain rule:
( x^2 + 3x ) ^2
(move exponent to front, recopy inside, multiply by derivative of inside):
2 (x^2 + 3x) * (2x + 3) Now use algebra to simplify.
Goodluck on the test Wednesday everybody!!!!!!!!!

Reflection #9 Terrio

This week we learned how to solve word problems in Calculus. I do not really understand this too well but I'll try my best to explain it. NOTES: Step one to solve Related Rates is identify all given quantities and quantities to be determined. Make a sketch and label the quantities. Step two is write and equation involving the variables whose rates of change either are given or are to be determined. Next, Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. And Last, substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change.

The problem I have is "the easiest part to solving these types of problems". I do not know how to identify all the given information, this is the first step so I can't even begin to start the word problems.

I can, however, take a derivative of an equation you might find in a problem like this...

1.x=2 dx/dt=5
2.y=x^2
3.1(dy/dt)=2x(dx/dt)
4.(dy/dt)=2(2)(5)
(dy/dt)=20

Reflection

This week in Calculus we learned how to solve word problems. Also we reviewed everything we have learned in Chapter 2, so this week i'll give a lot of review problems for my blog.

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.**

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

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

This is just a few of the many things we learned.

More Examples:

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)
= 0

Reflection

This week was mostly a review chapter two. In this chapter we learned many things we learned how to find a derivative the long way and the short cut way, to find out if something is continuous and differentiable, average velocity, the product rule and the quotient rule, how to find the equation of a tangent line, how to find the slope of a normal line, the chain rule, implicit derivative and explicit derivative, second derivative with implicit derivatives, and how to solve related rate problems. The one thing with this review that I’m still having trouble with is how to solve related rate problems.

The steps to solving these problems are:

1. identify all given quantites and quantities to be determined. Make a sketch and lable 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, implicitly 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.

Week 8 Blog Prompt

What are the steps to a related rate problem? What are the key words you look for and what do they mean? Give an example of a related rate problem and solve it.

Reflection #8 Terrio

This week we learned how to solve implicit derivatives :

Implicit functions means it's not solved for y and explicit functions means it is 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.

So when solving all you really do is take the derivative of the whole equation. After you take the derivative you just solve for dy / dx or y'. And 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 ).

Reflection #8

In honor of the quiz we took Friday, ( and due to the fact that I forgot my all my notes from this week in my locker ) I am going to do a reflection on the Product Rule and the Quotient Rule…..I know, dates back a few weeks but sorry. 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
-Hear is an example of the product rule:
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
Here is an example of the quotient rule:
X^2 + 3 / 2x
= ( x^2 + 3 ) ( 2 ) – ( 2x ) ( 2x ) / (2x)^2
= 2x^2 – 4x + 3 / (2x)^2

Lindsey's Reflection

This week we learned about implicit derivatives.

implicit means that is it not solved for y. while
an explicit function is solved for y.

whe you take a derivative you must put a notation by every variable
except x.

dy/dx means the the derivative of y
with respect to x. it can also
show up as y^1 which is y prime.

here are the steps to finding 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 a dy/dx

4. solve for dy/dx

5. simplify and if it is possible
to do so replace with the original
equation that was given.

here is an example


y^3+y^2-5y-x^2=-4

3y^2dy/dx+2ydy/dx-5dy/dx-2x=0

3y^2dy/dx+2ydy/dx-5dy/dx=2x

dy/dx(3y^2+2y-5)=2x

dy/dx= 2x/3y^2+2y-5

and that is the final
answer.
it sometimes is a long process
by ultimately it is very
easy to do.

Reflection

This week we learned how to solve problems when we have implicit derivatives, explicit derivatives and a second derivative in an implicit function. To solve for an implicit derivative and a explicit derivative you have to know the difference. An implicit derivative is not solved for y and an explicit derivative is solved for y. when your solving for either the implicit or explicit derivatives after you take the derivative of something with a y you have to put the dy/dx notation behind it.

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

Reflection

This week in calculus we learned section 2.5 and that involved implicit derivatives, explicit, and second derivative with implicit. 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

In section 2.6 we learned about related rates, and you must follow the guidelines from page 150 in the text book to solve these types of problems.

Example:

1) y = 4(x^2 - 5x)
1. dy/dt = ? x = 3 dx/dt = 2
2. y = 4(2x - 5)
3. dy/dt = 4(2x - 5)
4. dy/dt = 4(2(3) - 5)
= 4(6 - 5)
= 4

Week 7 Blog Prompt

What concept do you feel like you mastered this nine weeks? What concept did you struggle with the most? Why? What can you change this nine weeks in your study habits, etc to improve your grade?

Reflection #7

This week started off a little confusing because we were trying to learn the chain rule which cause me a little problems. I do understand how it works now and it is pretty easy. When you solve for the chain rule: you take the derivative of the outside, then derivative of the outside then recopy the inside, multiply by derivative inside.
d/dx f(g(x)) = f'(g(x)) X g'(x). When using the chain rule you still may have to use the product or quotient rule. product rule:(recopy first equation)(derivative of the second equation)-(recopy second equation)(derivative of first equation)
quotient rule:(bottom)(top derivative)-[(top)(bottom derivative)]/bottom equation squared
You will need these formulas and know some of the identities for the exam tomorrow. Remember that if the question says find the rules of differenation to find the derivative of the function x^4 +9x^5. All you have to do is find the derivative and because we learned the shortcut method, this is a very easy problem. The answer you get is 4x^3+45x^4. NOW STUDY FOR THE EXAM!!!!!!!!!!!!!!!!!!!!!

Reflection # 7

This past week was EXTREMELY confusing at first and to be honest I was really lost with the Chain Rule, but me and Chauvin sat down at his house one night and we had like an epiphany and weren’t lost anymore. So I will come explain how to do the chain rule because I fully understand it now! To start, you need to know that the chain rule, like everything else in calculus, has a formula to it. To begin the formula you need to figured out what “u” equals and what “y” equals. In simpler terms, figure out whats on the inside and whats on the outside. The you use the Chain Rule which says that you put:
Derivative of the outside (recopy inside) times the (derivative of the inside)
An Example:
Sq. root of ( x – 4 )
-recopy as ( x – 4 )^ ½
- y = u^1/4 and u = x – 4
-1/2 ( x – 4 ) * ( 1 ) (its 1 because the derivative of x – 4 is 1)
-so simplify that and you get x/2 + 2 and that’s the derivative of sq. root of ( x – 4 ) using the chain rule.
NOW EVERYONE STUDY FOR THE EXAM!!!!!!!!!!!!!!

This week we reviewed on how to solve for the chain rule and when you have a chain rule, quotient rule, and product rule inside a chain rule or when you have a chain rule inside of a chain rule, quotient rule, or product rule. When you solve for the chain rule you use the formula d/dx f(g(x)) = f'(g(x)) X g'(x). in word form you take the derivative of the outside, then derivative of the outside then recopy the inside, multiply by derivative inside. Here’s an example of when you have to figure out what rule is first.

Example: d/dx -7/(2t-3)^2
1. Quotient Rule
2. Chain Rule

In this problem you would use the quotient formula first then take the chain rule and plug that in.
(2t-3)^2(0) - [(-7)(2(2t-3)^1(2)]/(2t-3)^4
28(2t-3)/(2t-3)^4
28/(2t-3)^3

Here’s an example of how you use the chain rule.

Example: d/dx (x^2 + 1)^3

3u^2 (2x)
3(x^2+1)(2x)
=6x(x^2+1)^2

Lindsey's Reflection

This week we learned about the chain rule..
this is supposed to be the hardest
concept to learn in calculus because
there is not a set of operations to follow
when doing it. so you kinda
just have to know when to spot it

this is the most that you have to follow.

first you have to find the inside and outside

after you follow this

d/dx=f(g(x))=f^1(g(x))xg^1(x)

here it is in english: derivative of outside, recopy inside,
then multiply by derivative of inside


here is an example:

d/dx(x^2+1)^3

the outside is y=(u)^3

the inside is u= x^2+1

3(x^2+1)2x
=6x(x^2+1)^2

that is your final answer.

good luck to everyone on the exam.

Reflection #7 Terrio

This week was INTENSE!!!!!!!!!!!! Nah not really, but we did leard how to do the chain rule and for the most part it seems pretty easy. The formula for d / dx f (g ( x ) ) = f' (g ( x ) ) X g'( x ) or in words, which is a lotttttt easier to understand, ( derivative of outside ) ( plug in the inside ) times ( the derivative of the inside ). U stands for the inside of the problem and Y is the problem. Before you plug into the formula you first have to break down the problem and get the inside and outside.

Ex :

( cos x ) ^2

u= cos x

y= ( u ) ^2

-derivative of outside = 2( u )

-plug in inside = 2( cos x )

-multiply by derivative of inside = 2( cos x ) ( -sinx )

-your answer is 2cosx( -sinx )

Reflection 10/10

This week in calculus we did more on the chain rule and reviewed for the exam. The chain rule, in words, is you take the derivative of the outside, recopy the inside, then multiply by the derivative of the inside. "Outside"= y=f(u) "Inside"= u=...

Examples:

y= 1/x+1

outside= y=1/u
inside= u=x+1


y=sin(2x)

outside= y=sin(u)
inside= u=2x


y=sqrt3x^2-x+1

outside= y=sqrt(u)
inside= u=3x^2-x+1
__________________________________________________________

d/dx (x^2 + 1)^3

outside= y=u^3
inside= u=x^2 + 1


3u^2 (2x)
3(x^2+1)(2x)
=6x(x^2+1)^2

__________________________________________________________

Problem with the chain rule and the quotient rule. You have to find which rule is inside of the other.

Example:

d/dx -7/(2t-3)^2

1. Quotient Rule
2. Chain Rule


(2t-3)^2(0) - [(-7)(2(2t-3)^1(2)/(2t-3)^4

28(2t-3)/(2t-3)^4
28/(2t-3)^3
________________________________________________________

SO that pretty much sums up this week, EXAAAAAMMMM tomorrrroww! Study study study, lets all get A's.

Reflection

This week in calculus we learned about the chain rule and began reviewing for first nine week exams. When you solve for the chain rule: you take the derivative of the outside, then derivative of the outside then recopy the inside, multiply by derivative inside.
d/dx f(g(x)) = f'(g(x)) X g'(x)

Examples:

1) y = (4x-1)^3
y = u^2/3 u = 4x-1
3u^2 X 4
3(4x-1)^2 X 4
= 12(4x-1)^2

2) f(t) = (9t+2)^2/3
y = u^2/3 u = 9t+2
2/3u^-1/3 X 9
2/3(9t+2)^-1/3 X 9
6(9t+2)^-1/3
= 6/(9t+2)^-1/3

3) y = sin(pix)^2
cos(pix)^2 X 2pix
= 2pixcos(pix)^2

4) f(x) = cotx/sinx
(sinx)(csc^2x)-[(cotx)(cosx)]/(sinx)^2
sin^2x/sin^3x
1/sinx
= cscx

Also, just to review for the exam, we learned how to do derivative shortcuts, the limit process, discontinuity and much more.
Examples of derivative shortcut:

1) x^3
3x^2-1
= 3x^1

2) 4x^3 +2x -5
4(3)x^3-1 + 2 – 0
= 12x^2 + 2

3) x^7
7x^7-1
= 7x^6

Week 6 Prompt

Is there an order of operations for the chain rule? Why or Why not? Give several examples of chain rule problems.

Reflection #6

This week was a pretty busy week with homecoming, but I still managed to get my calc on. This week in calculus went pretty good. We had a test Friday that I hopefully did decent on. I knew most of the stuff, but the word problems caused me a little bit of confusion. If the whole test was find the derivative I would have had a party. Finding the derivative is extremely easy now. To solve it you do these formulas. The formula for a secant line is this m= f(c+deltax)- f(c)/ deltax. This is the secant slope lim f(x+deltax)/deltax deltax-->0. The three main steps to remember in doing this is expand cancel and factor out. We also learned to knew formulas, the product rule and quotient rule
-Product Rule:
( copy first equation ) ( derivative of the second equation ) - ( copy second equation ) ( derivative of first equation )
-Quotient Rule:
bottom ) ( top derivative ) - [( top ) ( bottom derivative )] / ( bottom equation squared )

Reflection # 6

This week I killed that test, I’m pumped haha. Anyway we pretty much jus reviewed and did the study guide all week so I will talk about some thing we learned the week before. The good ole product rule. When doin the product rule you just have to remember the formula which is:
( copy 1st ) ( derivative of 2nd ) + ( copy 2nd ) (derivative of 1st )
Here is an example:
( x^2 + 2 ) ( x – 3)
( x^2 + 2 ) ( 1 ) + ( x – 3 ) ( 2x)
X^2 + 2 + 2x^2 – 6x
= 3x^2 – 6x
The quotient rule works the same way, the only difference is the formula. …
( copy bottom ) ( derivative of top ) – ( copy top ) (derivative of bottom) / ( bottom ) ^2
I find the quotient rule is a little harder than the product rule just because the formula puts it in fraction form, but I works the same way.

Reflection #6 Terrio

This week was like a review week for us. We went over all the stuff we did in the past two weeks to get ready for the test on Friday. We did a packet that consisted of finding derivatives, product rule, quotient rule, finding velocity, and anything else we may have learned in the past two weeks.

-Finding the derivative is pretty easy now. If you are suppose to use the derivative formula all you do is plug in ( x- delta x ) everywhere you see an x then subtract by the original problem and put it all over delta x. Then simplify to get your answer. If it just says to find the derivative then you can use the shortcut: constants = 0. The derivative f( x ) = x^6 move the 6 to the front and subtract one from the exponent and your derivative would be 6 x^5...this is called the Power Rule.

-Product Rule and Quotient Rule:

-Product Rule:

( copy first equation ) ( derivative of the second equation ) - ( copy second equation ) ( derivative of first equation )

-Quotient Rule:

( bottom ) ( top derivative ) - [( top ) ( bottom derivative )] / ( bottom equation squared )

Lindsey's Reflection

this week we practiced, did our study guide, reviewed, and had a test.

we really did not learn anything new.

this included derivatives, and the product and quotient rule
all which dealt with formulas

derivative formula

lim (x+deltax)-f(x)/deltax
deltax-->0


product rule

(recopy first equation)(derivative of the second equation)-(recopy second equation)(derivative of first equation)

quotient rule

(bottom)(top derivative)-[(top)(bottom derivative)]/bottom equation squared


this is pretty much basically what we
covered this week.
nothing new and all easy
as long as you know the
formulas!
see everyone tomorrow.

This week it was really just a review of chapter 2 something we learned how to do in chapter 2 was use the product rule and the quotient rule. You use the product rule when something is being multiplied and the quotient rule when something is being divided. We also reviewed how to use the identities that we learned from last year to help us get the simplified answer.
Product Rule:f(x)g(x) = f(x)g’(x) + g(x)f’(x)(recopy 1st)(derivative 2nd) + (recopy 2nd)(derivative 1st)Example: g(x) = (x^2 + 3)(x^2 – 4x)g’(x) = (x^2 + 3)(2x – 4) + (x^2 – 4x)(2x)= 2x^3 – 12 +2x^3 – 8x^2= 4x^3 – 8x^2 – 12 / 4= x^3 -2x^2 -3
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)^2Example: f(x) = x / x^2 + 1f’(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. We also learned how to find derivatives and the average velocity of something. That was chapter 2.

Reflection

In calculus this week, we finished reviewing and learning about some of chapter 2. We also took a test on chapter 2. In this chapter we learned about derivatives. To help us with the derivatives, we learned how to use the shortcut, and we also learned different direction sets, plus we learned how to find average velocity and other things.

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:
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