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.
Sunday, October 18, 2009
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.
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 :)
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
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.
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.
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.
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?
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.
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.
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