Post #10

This week in calculus we learned related rates and angle of elevation.

The steps for related rates are:

1. Identify all your variables and equations. Some may be given, but some you may have to find.

2. Identify what you need to solve for.

3. Make a sketch or draw a picture of what you are given.

4. Take the derivative of the equations with respect to time.

5. Plug in given numbers to the derivative and solve.

EXAMPLE 1: The problem can be given as just and equation with given information such as :

x^2+y^2=25 dy/dt when x=3, y=4 dx/dt= 8

Derivative: 2x dx/dt + 2y dy/dt= 0

When you plug in you get: 2(3)(8) + 2(4) (dy/dt) = 0

48+ 8 dy/dt=0

dy/dt= -6

EXAMPLE 2: You may have to find the equation yourself:

The radius r of a circle is increasing at a rate of 3 centimeters per minute. Find the rates of change of the area when r=6 centimeters and when r=24 centimeters.

The area of a circle is A=pir^2

You are given dr/dt=3

Take the derivative of the equation: dA/dt=2pir (dr/dt)

Now plug into your derivative: dA/dt= 2pi(6)(3)

dA/dt= 36pi cm^2/min.

Using the same derivative, you plug in 24 into the equation instead of 6

dA/dt= 2 pi (24)(3)

dA/dt= 144 pi cm^2/min.

I understand the simple related rates problems such as these two; however, I don't understand how to work them when they get harder and give real life situations. The same thing applies for angle of elevation. I let the words get in the way and cannot find what they are asking for or what is given.

Help would be appreciated =)

10th post

This week in calculus was not to bad. In the beginning of the week we learned how to do implicit derivatives.
Steps:
1. Take the derivative
2. When you take the derivative of y, you put dy/dx behind it
3. Solve for dy/dx

An example of this would be:
6x^2+7y=9
(first take the derivative and put dy/dx behind the y derivative)

12x + 7 dy/dx=0
(now solve for dy/dx)

dy/dx= 12x/7

Another thing we learned this week is second implicit derivatives:
the steps are the same except after you solve for dy/dx, you are going to take the derivative of the function again and simplify it.

One thing i understood this week was related rates, its kind of like optimization but not as difficult lol. One way to simplify matters is to always write your given, then figure out what you are looking for. They may give you an equation and sometimes you may have to figure out one on your own. Once you have your equation, take the derivative in respect to time.

lets look at an example:

lets say they give you the radius r=6 and tell you r is increasing at a rate of 3 cm/sec.
You have to know the formula for area of a circle is A=(pie)r^2. Then they ask to find the rate of change of the area given these conditions.

First, take the derivative of the area formula

dA/dt= 2(pie)(6)(3)
dA/dt= 36(pie)
So your rate of change of area would be 36(pie), in related rates u basically plug into the formula, then take the derivative and solve for what they ask, in optimization your plugging into like 2 different things whereas related rates is mostly one formula and u look for one thing most of the time.

One thing i do not understand is angle elevation, i understand that it has the same concept as related rates considering the setup and the steps, i just do not understand how to find the information from looking at the triangle or solving the triangle. If someone can help me understand this it would help a lot. Thanks :)

Post #10

OHHH MMMYYYYYY....i don't know what to say about this week in calculus. At the beginning things were looking up! We learned about Implicit Derivatives, that's something i know how to do sooo...

steps to IMPLICIT DERIVATIVES:
1. Take the derivative like normal.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.

36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x

We then learned how to take the Second Implicit Derivative...

steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values

so an example using the same equation is:
we take the first implicit derivative FIRST...[durr]
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x <-- that's our first implicit derivative

now in order to take the second implicit derivative we just take the second derivative of dy/dx = -18x..

dy/dx = -18x
dy/dx = -18
so since we know that dy/dx is -18x we can plug that in
-18x = -18
when solving for x we get 1!

those, i understand...now for what i really do have trouble with is ALL of the word problems and the Angles!
If someone would take the time and go step by step and tell me how you know what's what it would be GREATLY appreciated!
once again, THANKS for the help!

10th Post

Calculus Week #10

Okay so this week we learned a few different things. We learned Implicit derivatives, related rates, and angle of elevation problems.

Implicit derivatives are terribly too easy. The only ones that even get remotely hard are the ones that have about 15 product rules, and even those are all about taking it slow and making sure to write the right thing down when you do them. The basic steps for implicit are just to take the derivative; however, whenever you take a derivative of a y term, make sure to write either y' (y prime which stands for derivative) or dy/dx (which stands for the derivative of y as well). Writing dy/dx is a better practice because similar notation comes in on related rates. Anyway, so you take the derivative, write dy/dx by the y terms you took a derivative and then you solve for dy/dx. So let's do an example.

x^2 + y^2 = 4

Derivative would be:

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

Now, solve for dy/dx to get:

dy/dx = -2x/2y

That would be your answer. :-) Now, second derivatives with implicit are the same exact thing. Except in your final answer, when you solve for d^2y/dx^2, you will have a dy/dx in your answer still. This is easily fixed, however, by plugging in the first derivative and then algebraically simplifying your answer.

So implicits are a joke. Anyway, related rates are fun too. You know you have a related rates problem when the problem seems to be like a broken record telling you "rates". Anyway, it's kind of like implicit, but with a picture. When you are doing related rates...you basically get a whole bunch of things...like the rate of this is blah, this is an equation that models another rate, at x=blah, what is the rate of blah2? Basically all you have to do, though, is take the derivative of the equation and plug in. Whenever you take the derivative of x or of y, since it is a related rates problem and the rates are all in reference to time, you do dx/dt, dy/dt. That basically says derivative of x in reference to time, or of y in reference to time. Anyway, then you solve for that it wants and plug in. If in the problem it gives you dx/dt already, just solve for dy/dt and then plug in to find your dy/dt. It's pretty simple once you get a hang of it. Sometimes the problems get a little more complex, but if you take your time and look at what's given and what it's asking for, it's really easy.

The last thing we got, which I'm slightly getting lost on, is the same thing as related rates really...sort of. It's angle of elevation. It always uses a triangle. And there is always one side that is staying constant. So, you take the derivative tan(theta)=y/x and plug in for y or x whichever one is constant. The only thing I get confused about here when they change up the problem too much. Like for instance, the one in homework about the shadow...I was a bit unsure about how that all worked out.

Can someone maybe explain to me how the shadow problem works out? (Yes, I did my homework for once lol. I'm glad I did too. :-D)

Anyway, overall good week. As soon as I get that problem I can go back to saying that this week was a breeze. :-) (so help me do that haha)

-John

Ash's 10th Post

Is it ONLY the 10th week of school? I feel as if I should be getting ready for finals! Not Halloween! v.v

This week was relatively easy! I actually got some of it! I couldn't come up with anything that looked right for the 2nd Implicit Derivatives, so I just took it from the notes. I might need a little bit of help with that if I ddin't get it right. I think I confused myself when I was typing it somehow.


Implicit Derivatives:
What you do for these is
1. Take the derivative like normal.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.

3x^2 + 4y = 9
1. 6x + 4(dy/dx) = 0
2. 4(dy/dx) = -6x
3. dy/dx = -6x/4


Second Implicit Derivatives:
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values

First Imp. Deriv:
x^2+y^2=25
2x + 2y(dy/dx) = 0
dy/dx = -x/y
Second Imp. Deriv:
d^2y/dx^2 = -[(y(1)-{(x)(dy/dx)}] / y^2
d^2y/dx^2 = (y-(x)(-x/y)) / y^2
d^2y/dx^2 = y+x^2/y/y^2/1
d^2y/dx^2 = (y^2 + x^2) / y^3
OR
d^2y/dx^2 = 25/y^3 (because (Y^2+x^2) = 25 in the orignal problem)

Now, as for what I don't understand:
Angle of Elevation!
When she's going over it, I can somewhat get it, but I don't think I'll ever be able to figure out what to do first second third etc. CAn someone help explain thise please? :) It's kinda like optimization: I knew how to do it somewhat when she was going over it, but I'm lost as a sigh in a windstorm when I'm trying to do it at home. I'll probably have homework questions so but I'm doing that tomorrow. :)

NEW POST

okay so since last week i was retarted and forgot my blog..im making sure this is done. therefore, this week we did a reviewed a little on implicite derivatives and learned about related rates and angle of elevation.

Implicite Derivatives guidelines:

1. involves Xs and Ys
2. a.)take the derivative like normal of both sides.
b.)everytime you take the derivative of y note it with dy/dx or y1.
c.)solve for dy/dx.
3. if you want the slope you must plug into a X and a Y value.

Example: y^3+y^2-5y-x^2=-4
3y^2dy/dx+2ydy/dx-5dy/dx-2x+0
dy/dx(3y^2+2y-5)+2x
dy/dx+2x/3y^2+2y-5

Related Rates:

1. identify all variables and equations.
2. identify what you are looking for.
3. make a sketch and label.
4. write an equation involving your variables. you can only have one unknown so a secondary equation may be given.
5. take the derivative with respect to time.
6. substitute in derivative and slope.

Example: the variables x and y are differentiable functions of t and are related by the equation y=2x^3-x+4, when x=2 dx/dt=1. find dy/dt when x=2.
dy/dt=?
x=2 dx/dt=-1
dy/dt=6x^2dx/dt-dx/dt
dy/dt=6(2)^2(-1)+(1)
dy/dx=-23

Related Rates:

1. Identify the known variables, including rates of change and the rate of change that is to be found.
2. Construct an equation relating the quantities whose rates of change are known to the quantity whose rate of change is to be found.
3. Differentiate both sides of the equation with respect to time (or other rate of change).
4. Substitute the known rates of change and the known quantities into the equation.
5. Solve for the wanted rate of change.

so i can honestly say i am having major problems with implicite derivatives and related rates. i really need help so if someone can explain it to me in a way that i can understand that would help so much.

Post...

I know it's late, but oh well.

This week after stressing for the exam, we learned something new that I actually got!!! Implicit Derivatives.

So you know it is an implicit derivative if it deals with both x's and y's. You also have to remember that when taking the derivative of a y term, you must either put a y' after or dy/dx. So example problems! Yay!

x^3 - 3x^2 + 2xy^2 = 12

After doing product rules and simplifying, you get:

3x^2 - (3x^2 y' + 6xy) + 4xyy' + 2y^2 = 0

solving for y' you get (-3x^2 + 6xy - 2y^2)/(-3x^2 + 4xy).

A listof things to remember:

1. y derivative dx/dy or y'
2. productrules
3. derivative of constant is 0
4. When possible factor out dx/dy or y'

Could someonehlep me on the t problems like the ones inthe mulitple choice practice.

This past week I was stressing out completly for the exam. I'm so thankful that we got to push that exam back, i probably would have failed other wise.

Thanks to john I finally learned how to optimize:

1. identify primary and secondary
(primary the one your maximizing or minimizing)(secodary the other one)

2. solve secondary for 1 variable and plug into primary

3.take derivative. plug into secondary equation.
- The problems were some of the last problems on chapter three study guide

I missed friday when we learned all of the new stuff,
anyone feel like taking the time to explain to me what
the heck we're doing and why we're doing it??

Post # 9

Okay, other than taking those exams... this week we learned explicit derivatives!!! Alright, to begin, Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx or y' (I prefer y').
3: Solve for dx/dy or y'
(if you want to find slope plug in an x and y value)

example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(y')+2y(y')-5(y')-2x=0

Then you have to solve for y', so you get:
y'(3y^2+2y-5)=2x which then is further solved for to get y'=2x/(3y^2+2y-5)

And that's all!!! this problem is finished.

Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)

First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(y'))=100(y+x(y'))

then, you need to foil it n stuff, so you get:
12x^3+12x^2(y')+12xy^2+12y^2(y'))=100y+100x(y')

then, as usual, you would have to solve for y':
y'=(-12^3-12xy^2+100y)/(12x^2+12y-100x)

after you solved for y', you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)

and I finally learned how to do optimization!!! I got that problem right on the free response part of the exam, but the thing I don't understand, was the second problem on that page. I forgot what it's called, and I don't remember it. All I know is that when I saw it, I didn't know what to do. If anyone remembers try to explain it please. Thanks.

Ash's 9th Post

HURRAY!! FIRST NINE WEEKS ARE OVER!!!! Ugh..that means we have three left? -.-
Anyway
This week:
1. Study, Study, Study
2. Study some more
3. Exam
4. Exam
5. Implicit Derivatives!

I totally get Implicit Derivatibes...so far...

First off: They invove x's and y's and have two sides: AKA: an equal sign...BUT DON'T FREAK!!
1. Take the Derivative of Both Sides
2. Take the Derivative of y just as you would x, except you put dy/dx or y' after the term...EVERY Term
3. Solve for dy/dy or y'..whichever you use

3x + 4y^2 = 3
1. 3 + 8y(dy/dx) = 0
3. 8y(dy/dx)=-3
y(dy/dx)=-3/8

It's totally simple so far.. :) Waiting for something to happen!

Hmm...what I don't get...does "I don't get people" count?
Nah..hmm..I honestly can't think right now..I think it's sleep deprivation.but I KNOW there's LOTS of things I don't get...Soon as I remember..I'll ask :)

Week 9

Finally, the first nine weeks are over. This week, we reviewed for our exam on Monday and Tuesday. On Wednesday, we took our multiple-choice exam, and on Thursday, we took our free response test. On Friday we learned something new: Implicit Derivatives.
I get Implicit Derivatives. They are basically just like regular derivatives, except they have a few extra steps. They must involve two constants (a x and a y) to be considered Implicit.
Steps:
Take the derivative like normal of both sides.
Every time you take the derivative of y – note it with dy/dx or y’.
Solve for dy/dx.
Example: y^2 + 5y = x^2
2y + 5 = 2x
Just took regular derivative like y was an x.
2y(y’) + 5(y’) = 2x
Notes y’ everywhere I took the derivative of y.
y’ = 2x/(2y+5)
Factored out y’ and divided (2y+5) over.
　
　
I also think I’m pretty good at finding the equation of a tangent line.
Steps:
Take f ‘(x).
Plug in x to find your slope: m.
Plug x into f(x) to find y (if not already given).
Using m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1).
I feel pretty confident, so I’m going to go out on a limb and say I don’t not understand anything that I can think of right now. But, I’m am getting ready for a long three MORE nine weeks.

Posting...#9

Finallly done the first nine weeks… That exam O.o I felt like I did really good on the multiple choice part but when it came to the free response I failed epicly but now I am going to get focused and do everything we get like I should have been so something like that doesn’t happen again to poor me L.
Well to start off the second nine weeks we learned implicit derivatives which involes x’s and y’s. To do implicit derivatives you

1. Take the derivative like normal of both sides
2. When you take the derivative of a y you not it with y’ or (dy/dx)
3. Solve for (dy/dx)
IF you want to find the slope you plug in the x and y value
An example of a implicit derivative is: y^3+y^2-5y-x^2=-4
1. 3y^2(dy/dx) +2y(dy/dx)-5(dy/dx)-2x=0
2. (dy/dx)(3y^2+2y-5)=2x
3. (dy/dx)=(2x)/(3y^2+2y-5)
Well that’s all we learned so far cause we only had one day so far. But when I was doing the home work this weekend I had a few problems. I had problems with all the most of the ones that involve like cos and sin and I really didn’t understand number 12, and number 8 the most because I have trouble when its like a #xy and I didn’t know how to worky around the pie symboly on number and for 21-28 I didn’t know how tho find the derivative through a specific point so if someone can explain that.
Im getting focused this nine weeks so I don’t bomb again.

Post Number Nine

This week in Calculus we continued reviewing for our exams up until Wednesday. That's when we took the multiple choice part, and thursday we took the free response. After doing all of the study guides multiple times i thought i was finally getting a hang of everything, but i probably thought wrong.

Friday, we started learning again. I was pretty nervous, but implicit derivatives seem to be pretty easy....so far. I'm still waiting on the catch though. The first thing you need to know about implicit derivatives is that it involves x's and y's. The steps are shown below:
1. Take the derivative like normal of both sides.
2. Every time you take the derivative of y note it with dy/dx or y'.
3. Solve for dy/dx or y'.

If you want the slope you must plug in a x and a y-value. Remember if you're only given an x and not a y, plug in the x value to the original function to find the y.

An example of an implicit derivative is y^3 + y^2 - 5y - x^2 = -4
The first thing you should notice is there is x's and y's, meaning you must take the implicit derivative.
3y^2 dy/dx + 2y dy/dx - 5 dy/dx - 2x = 0
dy/dx(3y^2 + 2y - 5) = 2x
dy/dx = 2x/3y^2 + 2y - 5.

Pretty simple, right :)

Some of the things i think i've finally gotten the hang of are tangent lines and limits.

Some things i am still having trouble with is of course, optimization, and looking at the graphs and being able to tell what the derivative looks like or where the points of inflection are, etc.

GOODNIGHT :)

post 9

This week in calculus we took our exam. We took the multiple choice portion on wednesday and the free response portion on thursday. We had six packets that we worked on for two weeks to help us review all the way up until the exam. it had chapters 1 through 3. It went over stuff from finding points of discontinuity of the graph of a limit approaching a number, all the way up to optimization. On Friday, we started learning again. We learned implicit derivatives. They are very easy, in fact, the only thing is you MUST know how to simplify correctly. It is just like finding a regular derivative, except instead of only having an x-value in the function, you have both an x-value and y-value. you find implicit derivatives by identifying whether or not it is an implicit der by seeing if it has a x and y value. you take the der of both sides. everytime you take the der of a y value you put dy over dx. then you solve for dy over dx

im still not completely comfortable with optimization because im not really good at finding the different variables. holla at me if you wanan help

post #9

This week in calculus we took our 1st nine weeks exam. We took the multiple choice portion on wednesday and the free response portion on thursday. We had six packets that we worked on for two weeks to help us review all the way up until the exam. It started with chapter 1 and went through chapter 3. It went over everything from finding points of discontinuity of the graph of a limit approaching a number, all the way up to optimization. By the way, after two weeks of confusion and completely failing a quiz, i finally understand optimization! it is so easy! On Friday, we started learning again. We learned implicit derivatives. They are very easy, in fact, the only thing is you MUST know how to simplify correctly. It is just like finding a regular derivative, except instead of only having an x-value in the function, you have both an x-value and y-value.

HOW TO FIND AN IMPLICIT DERIVATIVE:

1. identify whether or not it is an implicit derivative. (does it have an x and y value?)
2. take the derivative of both sides.
3. everytime you take the derivative of a y-value, you must put dy/dx behind it, or y'.
4. solve for dy/dx

Although, for implicit derivatives, sometimes it asks for the slope. If it does, it will give you a point (3,2) with an x & y value. If a point is not given, then an x-value is given, and it will say, find the slope of f(x) @ x=5. In this case, you would plug x into the original f(x) function and sovle for y to get your y-value.


I don't really have any questions this week, the only thing i get confused on is how to find points of inflection and max/mins.

Post #9

Exam week finally over! This week we learned how to take implicit derivatives which are derivatives when there are x and y values.  I think I understand the how to do them, but I get confused simplifying especially the last problem we did in class and I need to remember label my y derivatives.  Anyway the steps for solve implicit derivatives are: 

1.  Take the derivative like you would when taking a regular derivative.  All the same rules apply.

2.  Everytime you take the derivative of y you label it dy/dx or y'.

3.  Lastly, you solve of dy/dx or y' depending on what you noted it as.

EXAMPLE:

4y^3 + 2y^2 + 6y - x^2 = 5

Derivative:  12y^2 dy/dx + 4y dy/dx+ 6 dy/dx - 2x = 0

Add the 2x to have all dy/dx on one side

12y^2 dy/dx + 4y dy/dx + 6 dy/dx = 2x

Factor out a dy/dx

dy/dx ( 12y^2 + 4y + 6) = 2x

Finally you divide 2x by 12^2 + 4y + 6 to get your answer

2x/ 12y^2 + 4y +6

If you are trying to find the slope of a tangent line, it is the same steps as finding it with a regular derivative, except you use the implicit derivative steps to find the derivative.  You still, however, have to find a y-value by plugging in for x, take the derivative and set it equal to zero, and instead of solving for x, you solve of dy/dx and then plug in your x and y values and simplify if needed.  

For what I'm still having trouble with is looking at graphs of f'(x) and determining where the f''(x) is concave up or down or where f(x) is increasing or decreasing, such as  the questions on the short answer part of the.  How to find points of inflection given f'(x) also confuses me sometimes.  

I'm going attempt to do my homework now. 

post 9

This past week in calculus we took our first nine weeks exam. This took place on Wednesday and Thursday. Then we started to learn about implicit derivatives. This is not difficult at all as long as you remember how to take a derivative.



The steps to taking an implicit derivative are

1. Take the derivative of both sides as if you were just taking a regular derivative.

2. Note everytime you take the derivative of a y with dy/dx or y'

3. Solve for dy/dx by bringing it to one side and solving for it as it were an x.



Implicit derivatives can have similar questions to regular derivatives such as what is the slope of the equation or tangent line. If the problem asks for slope or tangent line and slope is needed but a point is not given but an x is, all you do is plug in the x given into the derivative to give you your x.



Also I learned that optimization is not as hard as I thought it was. Once I got a little help I started to understand it. And also I learned if you are looking for the dimensions of a rectangle to maximize the area it will be a square.

The only thing that confuses me is something like the last one we did together in class on Friday that we did not finish. When ever it gets complex like that it loses me and I just lose where im at.

Post 9

This week in calculus we had our exam. B-Rob gave us until Wednesday and Thursday to take our exam so we could have some extra time to do our packets and study. We did a lot of reviewing this past week to prepare us for the exam. We reviewed things like limits, derivatives, the first derivative test, the second derivative test, Rolle's theorem, mean value theorem, and ways to find either the origional function, first derivative, or second derivative from looking at a graph. Friday, after we were finished with our exams, we learned new material. This new material was implicit derivatives. They're just like regular derivatives, except there is more than one variable in the function.

The steps for finding implicit derivatives are as follows:

1. Take the derivative like normal of both sides

2. Every time you take the derivative of y, note it wigh dy/dx or y'

3. Bring all the dy/dx terms to one side and the regular ones to the other side of the equation

4. Solve for dy/dx

Also for implicit derivatives, if a problem is asking for the slope and it only gives you an x value, you plug the x value into the origional function to get a y.

Another thing to remember for implicit derivatives is if you see a problem that has x values on one side of the equal sign and y values on the other, leave them as they are and take the derivative and deal with them later.

Ok, this week my question is on implicit derivatives. I know the steps and I can follow them, but when I'm taking my derivative, I always forget to note dy/dx after I take the derivative of y or I get confused on where my dy/dx goes. If anyone knows a way to help me organize my work and a way to remember where to put dy/dx, please let me know.

Mambo Number Nine.

This week in Calculus I stressed, studied, hoped for the best, slightly cried after looking at the exam, took a deep breath, then started the exam. I hoped for the best, and it didn’t turn out so bad. The thing we learned new this week was implicit derivatives. This is really easy, its taking the derivative of a y.

Implict Derivative Steps:
1. take derivative of both sides (they have a equal sign and two variables (like x and y))
2. When you take the derivative of a y, you note it by dy/dx or y’
3. Then you solve for dy/dx or y’

Now, if you feel like being an overachiever, or the problem asks for it, the slope is found by…pluging in a x and y-value.

So, now I’m going to steal an example problem from class.

This is your equationy^3+y^2-5y-x^2=-4
Now, lets just simply take the derivative of both sides
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Now, simplify..notice I marked every time I took the derivative of y
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
now, begin solving for dy/dxdy/dx(3y^2+2y-5)=2x
still simplifyingggg..dy/dx=2x/3y^2+2y-5
TADAAA. ALL DONE :)

That’s really as good as it gets.

Just make sure that you don’t have to have a dy/dx ONLY behind y terms. For example, in product rule it might look like x(dy/dx) because you have to multiply the first term (x) times the derivative of the second term (y)..Which we know would be one and then noting that you took the derivative of y. Although this is really simple, it’s also really easy to make mistakes. So be sure you are working diligently. Hahah.

The thing I am confused on is finding the tangent line when you’re given the equation of like a circle or something random..it was on the take home tests and the exam..and I just don’t get it. I know how to do the normal way, and I know the steps, I just don’t understand when random decimals and points that make a circle and that nonsense. Anyone can help?

Post #9

This week we took our exam and learned about implict derivatives. Implict derivatives are not that hard. It is pretty much the things we learned like taking derivatives; it just has new steps.

Implict Derivative Steps:
1. take derivative of both sides (implict derivatives have an = sign)
2. every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx (you are going to have to take out a dy/dx when solving)
4. If you want the slope you must plug in a x and y-value.

Example:
y^3+y^2-5y-x^2=-4
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
dy/dx(3y^2+2y-5)=2x
dy/dx=2x/3y^2+2y-5

I pretty understand everything we have done except optimization. No matter how much I study the steps or do problems, optimization doesn't click. I just get stuck on different problems, and I forget little stuff like whether to plug into an original or derivative equation.