The third week of calculus was actually not bad at all. We started off the week going over derivatives that were going to be on the test. I was not understanding derivatives until tuesday. For some reason, something in my head just clicked and i understood what i needed to do. I was happy when i was able to work some hard derivative problems and get the right answers :). The only problem i still seem to have with derivatives is fully simplifying, but i'm sure i'll get better at it in due time. The test Wednesday was very challenging what with the firedrill and everything. But all in all i believe i did fairly well on the test. I thought the limits were going to hurt me, but i remembered the rules for limits.
1. If the degree of the top is larger than the degree on the bottom, the limit is infinity
2. If the degree on the top is the same as the degree on the bottom, you divide the coefficients
3. If the degree on the top is smaller than the degree on the bottom, then the answer is 0.
I also started to understand some of the concepts like removeables, you factor out the top and bottom of a fraction and anything that cancels is a removable. The derivative part of the test was better than i thought it would be lol.
On Thursday we started going over graphs of derivatives. I understand everything mrs. robinson went over on the graphs. How to know if its decreasing and increasing; and i understand about concave up and concave down. The only parts i am having problems with is when she gave us some equations and once we solved for x, we had to do a step that involved plugging back in for x, but we had to find different values than what were given. That's the only part that lost me a little bit, but i'm sure i'll get the hang of it.
So all in all, the third week of calculus was not bad. I was sure i was going to do very bad on derivatives, but i surprised myself lol. Even though we are learning new things now i'm sure i won't have any problems grasping the concepts this time. All i need is a few more examples and i'll be good to go. I'm looking forward to another good week!
Saturday, September 5, 2009
Friday, September 4, 2009
Post #3
Calculus :: thrid week
Soooo...we started off the week doing the same things we did last week, and i still don't all the way understand! But it's getting there i guess. We were still taking the derivatives of the BIG ones, it's just hard to figure out what to do first! Then we started with Arc Sin .. i actually understand a little better now! GO ME! Oh and we had our first TEST Wednesday! Wahoo...yeah that was HARD! it was on verticle asymptotes, horizontal symptotes, limits, points of discontinutity, and all of the derivatives we had learned so far. I studied alot! [even if it didn't pay off]
For verticle asymptotes you set the bottom equal to zero then after canceling you solve! For horizontal asymptotes it must be y = a number. To find it, take the limit of infinity, but remember if you get infity it DOES NOT EXIST! For example:
(x^2 + 4x + 9)
---------------
(4x^2 + x + 6)
since the HIGHEST exponet is the same then you divided the coefficients. Giving you y=one forth!
But, for
(x^3 + x)
----------
x
it's DNE because the greater one is at the top and it's infinity, which means no horizontal asyptotes!
Remember that for limits:
- if the degree at the top is greater there is no hoizontal asymptotes, and it's + or - infinity
- if the degree at the top is less then it's zero
- if the degree at the top and bottom are equal then you divide the leading coefficients to get your answer!
In order to find the limit of a number you can either plug it into the equation or, if you have a graph, just look at the graph!
To find points of discontinutity, always factop the top and bottom, then, if possible, make cancelations. Set the bottom equal to zero. If able to make cancelations, it's a removable ... for fractions, if you can still set the bottom equal to zero, then it's an asymptote, or infinite at that number! However if it's a piecewise then it could be a removable, jump, or it could be contiunuous. Solve by what is given. if it has an equal then it's not a removable. When you solve, if they're not the same it's a jump. But, if you solve and they are equal then it's continuous. Remember: if there's NO equal then it's a removable at that point!
Thursday and Friday we learned about ... increasin, decreasing, positive and negative slopes, concave up or down, horizontal tangents, derivative above the axis, as well as below the axis, what a zero of a derivative means, the max and min, and the points of inflection, knowing that the slope is set equal to zero! ... PHEWW ... you'd think it's hard by the way it sounds...but it's just basic concepts really!
We even learned about the first derivative test:
Soooo...we started off the week doing the same things we did last week, and i still don't all the way understand! But it's getting there i guess. We were still taking the derivatives of the BIG ones, it's just hard to figure out what to do first! Then we started with Arc Sin .. i actually understand a little better now! GO ME! Oh and we had our first TEST Wednesday! Wahoo...yeah that was HARD! it was on verticle asymptotes, horizontal symptotes, limits, points of discontinutity, and all of the derivatives we had learned so far. I studied alot! [even if it didn't pay off]
For verticle asymptotes you set the bottom equal to zero then after canceling you solve! For horizontal asymptotes it must be y = a number. To find it, take the limit of infinity, but remember if you get infity it DOES NOT EXIST! For example:
(x^2 + 4x + 9)
---------------
(4x^2 + x + 6)
since the HIGHEST exponet is the same then you divided the coefficients. Giving you y=one forth!
But, for
(x^3 + x)
----------
x
it's DNE because the greater one is at the top and it's infinity, which means no horizontal asyptotes!
Remember that for limits:
- if the degree at the top is greater there is no hoizontal asymptotes, and it's + or - infinity
- if the degree at the top is less then it's zero
- if the degree at the top and bottom are equal then you divide the leading coefficients to get your answer!
In order to find the limit of a number you can either plug it into the equation or, if you have a graph, just look at the graph!
To find points of discontinutity, always factop the top and bottom, then, if possible, make cancelations. Set the bottom equal to zero. If able to make cancelations, it's a removable ... for fractions, if you can still set the bottom equal to zero, then it's an asymptote, or infinite at that number! However if it's a piecewise then it could be a removable, jump, or it could be contiunuous. Solve by what is given. if it has an equal then it's not a removable. When you solve, if they're not the same it's a jump. But, if you solve and they are equal then it's continuous. Remember: if there's NO equal then it's a removable at that point!
Thursday and Friday we learned about ... increasin, decreasing, positive and negative slopes, concave up or down, horizontal tangents, derivative above the axis, as well as below the axis, what a zero of a derivative means, the max and min, and the points of inflection, knowing that the slope is set equal to zero! ... PHEWW ... you'd think it's hard by the way it sounds...but it's just basic concepts really!
- if the slope is positive it's increasing.
- if the slope is negative it's decreasing.
- above the x axis would be from zero to infinity.
- below the x axis would be from negative infinity to zero.
- if you're concave up then your slpe is positive.
- if you're concave down then your slpe is negative.
- a horizontal tangent is the slope of a horizontal line [which is zero].
- remember that concave up is like a bowl and concave down is like a frown.
- if increasing increasing or decresing decreasing occurs...it's neither the max or the min [just a reminder]
- finding the value of y ... just plug into the original!
We even learned about the first derivative test:
- take the derivative
- set equal to zero
- solve for x [which will give you max and min, hoizontal tangents, extrama..which is critical]
- set up intervals using the last step
- plug in the first derivative equation
- to find an absolute max/min: plug values from the last step into the original function.
- check endpoints
so yeah that's just about all folks!...have a great weekend! [and labor day]...
but...i still don't know what to do first for the BIG deriviatives!... :[
jessie's 3rd post:)
okay so third week of calculus has past by and it really wasnt as bad as what i thought it would be. i have to be honest and say that i have no idea what im going to write in this post. i thought that test was going to kill me and especially with that fire drill wasting everyones time.
so, this week was about review, test, and new things. using geogebra mrs robinson taught us about concave graphs and i actually got the concept pretty good it was alot easier than i thought.
i also learned that drawing things really helps solve a problem. aslo, if a line on a graph is going upward its positive and if its going down its negative which is pretty easy to understand. concave up is from that number to infinity concave down is that number to negative infinity. there is also a list to follow. and you need to know about horizontal and vertical lines.
i am still having trouble with some derivatives though i cant give an exact problem for someone to help me with but its usually larger one with alot of steps that mess me up. i try the breaking it down method with like listing if includes quotient rule and product rule etc. i am also still having problems cancelling things i dont need to. everytime i have to cancel something i end up double guessing myself and then i just do the wrong thing. so hopefully theres a way to help with that.
so, this week was about review, test, and new things. using geogebra mrs robinson taught us about concave graphs and i actually got the concept pretty good it was alot easier than i thought.
i also learned that drawing things really helps solve a problem. aslo, if a line on a graph is going upward its positive and if its going down its negative which is pretty easy to understand. concave up is from that number to infinity concave down is that number to negative infinity. there is also a list to follow. and you need to know about horizontal and vertical lines.
i am still having trouble with some derivatives though i cant give an exact problem for someone to help me with but its usually larger one with alot of steps that mess me up. i try the breaking it down method with like listing if includes quotient rule and product rule etc. i am also still having problems cancelling things i dont need to. everytime i have to cancel something i end up double guessing myself and then i just do the wrong thing. so hopefully theres a way to help with that.
Tuesday, September 1, 2009
the test
Sooooo, I just needed a break from studying and I'm a little nervous for this test.
just saying
just saying
FOR MY GRADE::
so i was wondering...if i can't really explain anything to anyone [or comment] because i don't understand it myself, can i just print out two of the one's i did last week! [not the same as the first time though]
Sunday, August 30, 2009
Post #2
So, my internet decides to work approximately 15 minutes before the blog is due (I've been trying for approximately 2 1/2 days). Lucky me! If it's late, don't blame me.
Ok, this week in Calculus was way harder for me than last week. I cannot simplify! Whereas last week was just the simpler stuff, this week we got into more complex, well, not concepts, but longer problems I guess you could say?
In the first week we learned how to take the quotient and product rules and how to take the derivatives of trig functions. For example, take the derivative of the following:
(x^2 - 2x + 4)/(x-4)
= (x-4)(2x-2) - (x^2 - 2x +4)(1) (quotient rule)
-----------------------------------
(x-4)^2
= (x^2 - 8x + 4)/(x-4)^2
This week, we also reviewed how to do logs before delving into the derivatives of logs. For instance:
Solve for x:
log39=x
All you do is switch it to so you have:
3^x = 9
and rather than taking the log of both sides like you would normally do, the answer is obviously x=2 because 3 squared equals nine.
We then went into taking the derivatives of natural logs, logs, and inverse trig functions. All I remember was that for the first night of homework (I think it was ln), I was really lost because even though I knew how to plug into the formulas, I just couldn't find out where to start or how to end it. But then I went to school and learned how to start. What really helped me was the list of things to do in the problem. And then I got a lesson in simplifying, and I got a little better. So, an example of inverse trig:
derivative of sin(arc cos(t))
= cos(arc cos(t)) * (-1)/(sqrt(1-t^2))
the cos and arccost cancel giving you:
t * (-1)/(sqrt(1-t^2))
which equals:
-t
--------------
sqrt(1-t^2)
For that particular problem, there wasn't much simplifying which is good for me. My only true problem is that I apparently cannot do basic algebra when simplifying.
And at the moment I'm going insane, so yeah, I think I'll end now.
Ok, this week in Calculus was way harder for me than last week. I cannot simplify! Whereas last week was just the simpler stuff, this week we got into more complex, well, not concepts, but longer problems I guess you could say?
In the first week we learned how to take the quotient and product rules and how to take the derivatives of trig functions. For example, take the derivative of the following:
(x^2 - 2x + 4)/(x-4)
= (x-4)(2x-2) - (x^2 - 2x +4)(1) (quotient rule)
-----------------------------------
(x-4)^2
= (x^2 - 8x + 4)/(x-4)^2
This week, we also reviewed how to do logs before delving into the derivatives of logs. For instance:
Solve for x:
log39=x
All you do is switch it to so you have:
3^x = 9
and rather than taking the log of both sides like you would normally do, the answer is obviously x=2 because 3 squared equals nine.
We then went into taking the derivatives of natural logs, logs, and inverse trig functions. All I remember was that for the first night of homework (I think it was ln), I was really lost because even though I knew how to plug into the formulas, I just couldn't find out where to start or how to end it. But then I went to school and learned how to start. What really helped me was the list of things to do in the problem. And then I got a lesson in simplifying, and I got a little better. So, an example of inverse trig:
derivative of sin(arc cos(t))
= cos(arc cos(t)) * (-1)/(sqrt(1-t^2))
the cos and arccost cancel giving you:
t * (-1)/(sqrt(1-t^2))
which equals:
-t
--------------
sqrt(1-t^2)
For that particular problem, there wasn't much simplifying which is good for me. My only true problem is that I apparently cannot do basic algebra when simplifying.
And at the moment I'm going insane, so yeah, I think I'll end now.
Post #2
This week in calculus we learned about line tangents, we did more advanced derivatives, and we also condensed logs. I didn’t have any problems with solving the logs but I kind of forgot how to condense logs. I can’t belive I forgot how to do logs. O.o I’m pretty sure I grasp the arc equations such as: [arcsin], [arctan], and etc., but I still have trouble with derivatives kind of because I always stop to earlier or mess up on my algerbra but that’s something I have to practice. I also don’t understand how to use (e) fully. Some problems that will really help me out if:
Just condense the problem, #32 last week’s homework.1) 2[lnx-ln(x+1)-ln(x-1)] and
Find an equation of the tangent line #50 last week’s homework.2) y= e-2x+x^2, (2, 1) .
Other than everything we learned; this week was pretty easy if you ask me.
Well I have to do a lot of studying and extra work but with a few examples like this I should be able to learn from what I know and what everyone helps me out with.
I don’t know if I’m supposed to be asking about this because it’s not from last week but for vertical and horizontal asymptotes . I understand vertical asymptotes but I don’t know how to get horizontal.
Like for:
3/x^2-4 you get 3/(x-2)(x+2)
Which gives you vertical asymptotes at x=2, and x=-2but I don’t know how to find the Horizontal ones
Thanks I hope I was able to ask that.
Mabile
Just condense the problem, #32 last week’s homework.1) 2[lnx-ln(x+1)-ln(x-1)] and
Find an equation of the tangent line #50 last week’s homework.2) y= e-2x+x^2, (2, 1) .
Other than everything we learned; this week was pretty easy if you ask me.
Well I have to do a lot of studying and extra work but with a few examples like this I should be able to learn from what I know and what everyone helps me out with.
I don’t know if I’m supposed to be asking about this because it’s not from last week but for vertical and horizontal asymptotes . I understand vertical asymptotes but I don’t know how to get horizontal.
Like for:
3/x^2-4 you get 3/(x-2)(x+2)
Which gives you vertical asymptotes at x=2, and x=-2but I don’t know how to find the Horizontal ones
Thanks I hope I was able to ask that.
Mabile
Post 2
So this week in calculus was intense. It seemed we were going at such a fast pase this week. For the most part, I grasped the concept of things. I understood the very beginning of everything, but got lost during the way. I'm not sure if there's anything I fully understand in calculus from this week, but I'll try to recall something I understood more than other things.
On Monday, we defined a derivitave again, but for different reasons. We learned how to find the equation of a tangent line.
Come to think of it, this is the only new thing we really learned this week, ha. Anyway, I pretty much understood this, so I'll explain it to the best of my abilities
When looking for the equation of a tangent line, once you get your x value, y value, and slope, you are to plug all thoes in to your point slope formula. To get these values, you are to do these steps:
The problem will give you an equation
Ex: f(x)=(x - 1)(x^2 - 2)
The problem will also give you a point
Ex: (0, 2)
1. The first thing to do is to find your slope. You will need to take the derivitave of the equation
Ex: (x - 1)(2x) + (x^2 - 2)(1)
When the derivitave is simplified, it will look like
3x^2 - 2x - 2
2. You then plug in your x value into your x, giving you your slope.
Ex: 3(0)^2 - 2(0) - 2
this will leave you with (-2) being your slope.
3. You thel plug in all your values into point slope form (y - y1 = m(x - x1)
When the x value, the y value, and the slope are plugged into this form, you will get
y - 2 = -2(x)
I also understood most of the log stuff.
I still have a hard time simplifying things. I see things that look alike and I try to combine them, when they can't be combined, or I don't combine things that can be combined. I think I just need more practice on it.
On Monday, we defined a derivitave again, but for different reasons. We learned how to find the equation of a tangent line.
Come to think of it, this is the only new thing we really learned this week, ha. Anyway, I pretty much understood this, so I'll explain it to the best of my abilities
When looking for the equation of a tangent line, once you get your x value, y value, and slope, you are to plug all thoes in to your point slope formula. To get these values, you are to do these steps:
The problem will give you an equation
Ex: f(x)=(x - 1)(x^2 - 2)
The problem will also give you a point
Ex: (0, 2)
1. The first thing to do is to find your slope. You will need to take the derivitave of the equation
Ex: (x - 1)(2x) + (x^2 - 2)(1)
When the derivitave is simplified, it will look like
3x^2 - 2x - 2
2. You then plug in your x value into your x, giving you your slope.
Ex: 3(0)^2 - 2(0) - 2
this will leave you with (-2) being your slope.
3. You thel plug in all your values into point slope form (y - y1 = m(x - x1)
When the x value, the y value, and the slope are plugged into this form, you will get
y - 2 = -2(x)
I also understood most of the log stuff.
I still have a hard time simplifying things. I see things that look alike and I try to combine them, when they can't be combined, or I don't combine things that can be combined. I think I just need more practice on it.
week 2
hellllooo claculus.
so to start off on a good note, i atleast get half of the stuff we did this week =)
first of all, i spent all of last year failing at life on our advanced math
tests because i couldn't grasp the concepts of logs if my life depended
on it, but it's a new year babyyy. when i first looked at our work sheet
i wanted to qluit life, but then it all clicked and i was super excited,
which i still am if you can't tell. though i am a little aggravated that
i spent so much time sucking at such simple concepts.
Now when i see someting like
evaluate the expression log279
i actually know what to do with it and not just put a ridiculous answer.
also, the equation of a tangent line, thankfully,
was extremely easy for me. i understood from the first example
Recall :
• A Tangent Line is a line which locally touches
a curve at one and only one point.
• The slope-intercept formula for a line is y = mx + b,
where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).
This formula uses a point on the line, denoted by (x1, y1),
and the slope of the line, denoted by m, to
calculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangent
line to a curve at an indicated point.
The equation for the slope of the tangent line to
f(x) = x2 is f '(x), the derivative of f(x).
f(x) = x2
f '(x) = 2x (1)
Therefore, at x = 2, the slope of the tangent line is f '(2).
f '(2) = 2(2)
= 4
Now , you know the slope of the tangent line, which is 4.
All that you need now is a point on the tangent line to be
able to formulate the equation. To find that point, simply plug
the coordinate of the shared point into the original equation, this gives you (2,4)
The only step left is to use the point (2, 4) and slope, 4,
in the point-slope formula for a line. Therefore: Y-4=4(x-2)
NOW FOR WHAT I DON'T UNDERSTAND.
.. they whole arc trig formulas really throw me off and i keep getting
confused on what exactly i do first, or what part of the problem i need
to simplify first, but since b-rob posted that response earlier i
think i understand a little bit better, i just need to stay motivated
and make myself work more problems becuase i know that is
the only way i'm fully going to understand and remember all this.
so to start off on a good note, i atleast get half of the stuff we did this week =)
first of all, i spent all of last year failing at life on our advanced math
tests because i couldn't grasp the concepts of logs if my life depended
on it, but it's a new year babyyy. when i first looked at our work sheet
i wanted to qluit life, but then it all clicked and i was super excited,
which i still am if you can't tell. though i am a little aggravated that
i spent so much time sucking at such simple concepts.
Now when i see someting like
evaluate the expression log279
i actually know what to do with it and not just put a ridiculous answer.
also, the equation of a tangent line, thankfully,
was extremely easy for me. i understood from the first example
Recall :
• A Tangent Line is a line which locally touches
a curve at one and only one point.
• The slope-intercept formula for a line is y = mx + b,
where m is the slope of the line and b is the y-intercept.
• The point-slope formula for a line is y – y1 = m (x – x1).
This formula uses a point on the line, denoted by (x1, y1),
and the slope of the line, denoted by m, to
calculate the slope-intercept formula for the line.
• The first derivative is an equation for the slope of a tangent
line to a curve at an indicated point.
The equation for the slope of the tangent line to
f(x) = x2 is f '(x), the derivative of f(x).
f(x) = x2
f '(x) = 2x (1)
Therefore, at x = 2, the slope of the tangent line is f '(2).
f '(2) = 2(2)
= 4
Now , you know the slope of the tangent line, which is 4.
All that you need now is a point on the tangent line to be
able to formulate the equation. To find that point, simply plug
the coordinate of the shared point into the original equation, this gives you (2,4)
The only step left is to use the point (2, 4) and slope, 4,
in the point-slope formula for a line. Therefore: Y-4=4(x-2)
NOW FOR WHAT I DON'T UNDERSTAND.
.. they whole arc trig formulas really throw me off and i keep getting
confused on what exactly i do first, or what part of the problem i need
to simplify first, but since b-rob posted that response earlier i
think i understand a little bit better, i just need to stay motivated
and make myself work more problems becuase i know that is
the only way i'm fully going to understand and remember all this.
"pzost 2" haha
The second week of Calculus AB was harder than the first week, although it wasn’t too bad. It really was a continuation of last week’s materials, but we just dug deeper into the information and the problems got harder. In the beginning of the week, we did more advanced product and quotient rule problems. Then, towards the end of the week we learned the derivative rules for arc trig functions.
I understood it for the most part, I just still have some problems with simplifying. Mrs. Robinson practiced a lot with us on it this week, so I don’t think I will too much more trouble with it and it should only get easier. I wasn’t really confused with anything this week because it was just more advanced than last week. So I’ll just explain how to take the derivative of the arc trig functions.
The formula for arccos and arcsin are the same, (1/√1-u^2)(u’). The formulas for arctan and arccot are also the same, (1/1+u^2)(u’), the only difference is that arccot has a negative in front. For example, if you had arctan(3x^2 + 4x) you would plug it into the arctan or tan^-1 formula. (1/1+(3x^2 + 4x)^2 (6x+4). Then you would simplify it. (6x+4/1+9x^4+24x^3+16x^2). These formulas are not hard to understand, it’s just very easy to make simple mistakes. The only thing I’m confused about still is the simplifying, and when to or when not to take the derivative of something while you’re simplifying. I know you are supposed to work from the outside in whenever you are taking derivatives, and that did help me a lot this week. Just need to keep practicing J
I understood it for the most part, I just still have some problems with simplifying. Mrs. Robinson practiced a lot with us on it this week, so I don’t think I will too much more trouble with it and it should only get easier. I wasn’t really confused with anything this week because it was just more advanced than last week. So I’ll just explain how to take the derivative of the arc trig functions.
The formula for arccos and arcsin are the same, (1/√1-u^2)(u’). The formulas for arctan and arccot are also the same, (1/1+u^2)(u’), the only difference is that arccot has a negative in front. For example, if you had arctan(3x^2 + 4x) you would plug it into the arctan or tan^-1 formula. (1/1+(3x^2 + 4x)^2 (6x+4). Then you would simplify it. (6x+4/1+9x^4+24x^3+16x^2). These formulas are not hard to understand, it’s just very easy to make simple mistakes. The only thing I’m confused about still is the simplifying, and when to or when not to take the derivative of something while you’re simplifying. I know you are supposed to work from the outside in whenever you are taking derivatives, and that did help me a lot this week. Just need to keep practicing J
Ash's 2nd post
Okay, well. I'm pretty sure the second week in Calculus was a trillion times harder than the first. Algebra. I think I need to re-take that class. Or at least a class in patience. Either one will work.
So, before I get myself into a mess, let me clarify something: I'm having a hard time remembering what we did this week and HOW to do it, so if I screw up, I blame my memory.
I think the safest thing for me to explain is the Trig Inverses. I half-way like them. Except the fractions.
Let's try an example problem! Arctan(3x^2+4x)
1. Find and state the formula for Arctan.
1
--------- * u'
1+u^2
2. Replace u for what is inside of the parenthesis.
1
--------------- * (3x^2+4x)'
1+(3x^2+4x)^2
3. Take the derivative of what you are multiplying it by.
(3x^2+4x) = (6x+4)
4. So now you would get
1
--------------- * (6x+4)
1+(3x^2+4x)^2
5. Since (6x+4) = (6x+4)/1, when you multiply, you will get this
(6x+4)
---------------
1+(3x^2+4x)^2
6. Now, foil out (3x^2+4x)^2
(6x+4)
------------------------
1+9x^4 + 24x^3 + 16x^2
7. Simplify what you can and you're finished!
2(3x+2)
------------------------
9x^4 + 24x^3 + 16x^2 + 1
Let's try another half-way simple problem.
Arccos(1/x^2)
1. Find and state the formula for Arccos. (Sense a pattern here?)
-1
----------- * u'
sqrt(1-u^2)
2. Replace u with what is in the parenthesis.
-1
----------------- * (1/x^2)'
sqrt(1-(1/x^2)^2)
3. Take the derivative of what you're multiplying by.
(1/x^2) = (-2/x^3)
4. Multiply to get
2
-------------------
x^3(sqrt(1-(1/x^4))
5. Simplify (completely simplified already) and you're finished!!
I don't know if that was clear or not. I hope so. If someone finds a mistake, let me know please. :)
As for my question...
I completely forgot logs (figures, used to be my favorite).
Can someone explain how to do this problem, I think it's the hardest:
ln ((x^2-1)/x^3)^3
Logs + Fractions + Derivatives /=/ Happiness.
Thanks!!
So, before I get myself into a mess, let me clarify something: I'm having a hard time remembering what we did this week and HOW to do it, so if I screw up, I blame my memory.
I think the safest thing for me to explain is the Trig Inverses. I half-way like them. Except the fractions.
Let's try an example problem! Arctan(3x^2+4x)
1. Find and state the formula for Arctan.
1
--------- * u'
1+u^2
2. Replace u for what is inside of the parenthesis.
1
--------------- * (3x^2+4x)'
1+(3x^2+4x)^2
3. Take the derivative of what you are multiplying it by.
(3x^2+4x) = (6x+4)
4. So now you would get
1
--------------- * (6x+4)
1+(3x^2+4x)^2
5. Since (6x+4) = (6x+4)/1, when you multiply, you will get this
(6x+4)
---------------
1+(3x^2+4x)^2
6. Now, foil out (3x^2+4x)^2
(6x+4)
------------------------
1+9x^4 + 24x^3 + 16x^2
7. Simplify what you can and you're finished!
2(3x+2)
------------------------
9x^4 + 24x^3 + 16x^2 + 1
Let's try another half-way simple problem.
Arccos(1/x^2)
1. Find and state the formula for Arccos. (Sense a pattern here?)
-1
----------- * u'
sqrt(1-u^2)
2. Replace u with what is in the parenthesis.
-1
----------------- * (1/x^2)'
sqrt(1-(1/x^2)^2)
3. Take the derivative of what you're multiplying by.
(1/x^2) = (-2/x^3)
4. Multiply to get
2
-------------------
x^3(sqrt(1-(1/x^4))
5. Simplify (completely simplified already) and you're finished!!
I don't know if that was clear or not. I hope so. If someone finds a mistake, let me know please. :)
As for my question...
I completely forgot logs (figures, used to be my favorite).
Can someone explain how to do this problem, I think it's the hardest:
ln ((x^2-1)/x^3)^3
Logs + Fractions + Derivatives /=/ Happiness.
Thanks!!
post 2
Well another week down in calculus and once again I understand some things but I don't understand some things. One thing i understand is get the equation of tangent lines. All you do is take a point and plug it into the derivative of the formula. But if only an x is given then you plug the x into the formula to get your y. And finally you have to put the slope you get when you plug it in the derivative and point in point slope form.
An example is x^3+2x^2-x+1 (2,-1). Then the derivative is 3x^2+4x-1. Now you plug the point in to get the slope. 3(2)^2+4(2)-1. This equals 19 which is your slope. Finally all you do is plug your point and slope into point slope form which is y+1=19(x-2). And even thought this can be condensed into a simpler form if you do it would not be in point slope form. So when you get something like that just leave it.
Also I understand how to get the equation of a normal line. You do the same things as you do for tangent lines except you just make the slope perpendicular. Using the same example above the answer would be y+1=-1/19(x-2). All you do to get the inverse of the slope is take the negative reciprocal.
But on the other hand, I don't understand how to do problems such as ln(lnx^2) or ln(t)/t. I don't even know how to start on these. If someone can get me started and explain it to me it would be greatly appreciated.
An example is x^3+2x^2-x+1 (2,-1). Then the derivative is 3x^2+4x-1. Now you plug the point in to get the slope. 3(2)^2+4(2)-1. This equals 19 which is your slope. Finally all you do is plug your point and slope into point slope form which is y+1=19(x-2). And even thought this can be condensed into a simpler form if you do it would not be in point slope form. So when you get something like that just leave it.
Also I understand how to get the equation of a normal line. You do the same things as you do for tangent lines except you just make the slope perpendicular. Using the same example above the answer would be y+1=-1/19(x-2). All you do to get the inverse of the slope is take the negative reciprocal.
But on the other hand, I don't understand how to do problems such as ln(lnx^2) or ln(t)/t. I don't even know how to start on these. If someone can get me started and explain it to me it would be greatly appreciated.
Order of Operations Attempt
Ok. This will not work for EVERY problem you encounter but it will help I think.
1. Product/Quotient Rule (if it is not inside parenthesis)
2. Exponent Rule
3. Trig Rules/log rules, trig inverse rules, e rules
You work from the outside in. So if the quotient is inside a log you would skip to three because it would be inside parenthesis. Or if a fraction is raised to an exponent then you would default to step 2 because it is inside parenthesis. I hope this helps!
pzost # 2
This week in class i managed to pick up somewhat youseful information that i will be using in calculus. I learned arc trig functions. we learned a total of six functions. These functions include sin, cos, tan, sec, csc, and cot. The function that I understand the sin function the most. The sin trig function is one over the square root of 1 -U2 times U prime. Even though i understand the concept of this function after i set the problem up and work it the first part ( plugging in and finding the derivitive, the first simple simplify) I get lost with the algebra. I have the very very very very very bad bad bad habbit of wanting to cancel out what i cant. I also get very very confused when i have the big massive fraction with in like two other fractions that i am suppose to multiply with some beast derivitive. And yes i need help wwith this.
Post #2
Alright, so...this week was extremely tough. I think most people probably agree on that, haha. We learned about finding equations for a tangent line, normal line, the points where y has horizontal tangents, and we worked on arc trig derivative formulas. I'm having so much trouble with solving the problem after I get the derivative, and I don't know why.
At least I get one thing we learned. It was finding the equation of the tangent line.
For example: (x^3-3x-1)(x+2) and given the point (1,-3)
First you just follow the product rule for the derivative: (x^3-3x-1)(1)+(x+2)(3x^2-3)
= (x^3-3x-1)+(3x^4-3x+6x^2-6) = 3x^4+x^3+6x^2-6x-7
Next plug in your x, which is 1: 3(1)^4+(1)^3+6(1)^2-6(1)-7 = 3+6-6-7 = -4
Finally, plug in to the point slope form: y+3=-4(x-1)
Something I am confused on is the horizontal tangents.
For example: y=3x^3+4x^2+5 at (1,12) So I understand how to get 9x^2+8x=0
But then what's next? How do I get the points?
Also, can someone help me on how to solve: arcsin3x/x ?
That would be the quotient rule. In the second part of the formula I'm suppose to copy the top, but I'm not suppose to have trig in the problem. So, do I take the derivative of it anyway? Or is there something else I'm suppose to do?
Overall, my confusion is mainly on simplifying these arc trig formulas once I have the derivative. They just seem massive and complicated. And then I get aggravated with it and give up. Hopefully, I get it by the test ha.
At least I get one thing we learned. It was finding the equation of the tangent line.
For example: (x^3-3x-1)(x+2) and given the point (1,-3)
First you just follow the product rule for the derivative: (x^3-3x-1)(1)+(x+2)(3x^2-3)
= (x^3-3x-1)+(3x^4-3x+6x^2-6) = 3x^4+x^3+6x^2-6x-7
Next plug in your x, which is 1: 3(1)^4+(1)^3+6(1)^2-6(1)-7 = 3+6-6-7 = -4
Finally, plug in to the point slope form: y+3=-4(x-1)
Something I am confused on is the horizontal tangents.
For example: y=3x^3+4x^2+5 at (1,12) So I understand how to get 9x^2+8x=0
But then what's next? How do I get the points?
Also, can someone help me on how to solve: arcsin3x/x ?
That would be the quotient rule. In the second part of the formula I'm suppose to copy the top, but I'm not suppose to have trig in the problem. So, do I take the derivative of it anyway? Or is there something else I'm suppose to do?
Overall, my confusion is mainly on simplifying these arc trig formulas once I have the derivative. They just seem massive and complicated. And then I get aggravated with it and give up. Hopefully, I get it by the test ha.
Post #2
I struggled this week. I’m not completely comfortable with anything we did this week but I think I understood how to find the equation of a tangent line over everything else. To find an equation of a tangent line, you can be giving a point or only giving an x value. If you are only giving an x value, you have to plug in x to find your y and make a point. Next you have to find the derivative of the equation. Once you found the derivative, you plug in for x and get your slope. The last step is to put in point slope form which is y-y1=m(x-x1).
An example of this is:
3x^3+4x^2=5 at (1,12) – you are already giving the point so you can skip step one
Derivative = 9x^2+8x
Plug in x = 9(1)^2+8(1)= 17
Plug into point slope form: y-12=(x-1)
I understand logs from last year and the first column of our page three worksheet but I didn’t understand the derivatives of logs. I know the formula is u prime/ulna but I get confused on the really big problems when you have to do more than one step to solve it. Making a list sort of helps but I end up doing it in the wrong order and my answer usually turns out completely wrong. If I do somehow happen to get the derivative right, I get stuck on the algebra part of the equation. I think I just need more practice on derivatives all together, especially before the test Wednesday. Help if you can and good luck on the test.
An example of this is:
3x^3+4x^2=5 at (1,12) – you are already giving the point so you can skip step one
Derivative = 9x^2+8x
Plug in x = 9(1)^2+8(1)= 17
Plug into point slope form: y-12=(x-1)
I understand logs from last year and the first column of our page three worksheet but I didn’t understand the derivatives of logs. I know the formula is u prime/ulna but I get confused on the really big problems when you have to do more than one step to solve it. Making a list sort of helps but I end up doing it in the wrong order and my answer usually turns out completely wrong. If I do somehow happen to get the derivative right, I get stuck on the algebra part of the equation. I think I just need more practice on derivatives all together, especially before the test Wednesday. Help if you can and good luck on the test.
Post 2
the second week of calc is over and i feel like i know a good bit even though its eof my hardest classes.it's getting better. I am understanding everything a lot better.
last year i slacked during the log chaper and i didnt know it because i played block dude everyday. I get everything from last year with the normal logs too. john taught me all the logs from last year. Log4=x which equals x=2 thats prettty easyyy. i understand the natural log properties from this past week when you plug everthing into the formula and simplify. Its pretty easy most times. we have ln(x^5). You plug into the ln formula, 1/u * u', and you get 1/x^5 * 5x. When you multiply you get 5x/x^5 which simplifies to 5/x (you cancel the x's). I understand that you may have to deal with multiple formulas in one problem. An example would be ln(x/(x^5 + 1)). In this problem you can see that you have to deal with natural log and quotient rule.
Something I don't understand is the difference from u^n and a^u. and i dont understand the arc trig functions with all the algebra. all the algebra in these problems mess me up and i end up messing it up. i hear sarah is pretty good at those btw hahahaha
everyone is at my house at the moment studying the test we have on wednesday. is that a test or a quiz because we are not too sure. can someone answer that question as well?
last year i slacked during the log chaper and i didnt know it because i played block dude everyday. I get everything from last year with the normal logs too. john taught me all the logs from last year. Log
Something I don't understand is the difference from u^n and a^u. and i dont understand the arc trig functions with all the algebra. all the algebra in these problems mess me up and i end up messing it up. i hear sarah is pretty good at those btw hahahaha
everyone is at my house at the moment studying the test we have on wednesday. is that a test or a quiz because we are not too sure. can someone answer that question as well?
Post Number Two
So everyone, the second week of calculus is over. Of course it's gotten harder, but not everything in life comes easily right? I felt good at the beginning of the week because we were reviewing quotient rule and product rule. Though at the time of learning them I thought those were impossible, I'm now wishing that we were back to that again. I'm confused about everything overall, but there are some things that I did understand this week (Thank God).
For example, I caught on to horizontal tangent points real quick. All you do is find the derivative of the problem given then set equal to zero and solve for x. We were drilled on solving for x last year, so I felt extremely comfortable doing that. An example problem of this is Find the points where y has horizontal tangents y=3x^3 + 4x^2 + 5. Therefore, y'=9x^2 + 8x because you just take the simple derivative of a polynomial. Once finding the derivative, set equal to zero giving you 9x^2 + 8x = 0. Factor out an x giving you x(9x + 8) = 0. You then find that your x points are 0 and -8/9. Next, you must plug in your x's to the original equation in order to find your ys. 3(0)^3 + 4(0)^2 + 5 and 3(-8/9)^3 + 4(-8/9)^2 + 5. These come out to give you the points (0,5) and (-8/9, 6.053). Box it off and that's your answer :)
Also, i understood the arc problems until we got to problems like #53 y=1/2(1/2ln x+1/x-1 + arctanx). I'm not exactly sure what you do first because there's a natural log and an arc formula. If anyone can direct me on this it'd be greatly appreciated.
I still remember the log properties from advanced math so I do okay on those problems. I also do fine on natural log problems unless they have more than one function in it. Like I said I get extremely lost then and have no idea where to start. Basically, I'm doing fine in this class except when it gets to the simplifying parts. I know it's just algebra but I just can't seem to grasp how to do all the right cancellations. I'm sure it'll take a lot of practice which I'm working on but it's just so frustrating not getting it time after time.
Finally, I want to extend the offer to arrange a study group before Wednesday to prepare for the test. I can always use a lot of help so if anyone is interested please let me know because I believe we can all help each other if we work together.
Overall, i'm looking forward to another stressful week full of work and headaches in Calculus--SIKEEEEEEE.
For example, I caught on to horizontal tangent points real quick. All you do is find the derivative of the problem given then set equal to zero and solve for x. We were drilled on solving for x last year, so I felt extremely comfortable doing that. An example problem of this is Find the points where y has horizontal tangents y=3x^3 + 4x^2 + 5. Therefore, y'=9x^2 + 8x because you just take the simple derivative of a polynomial. Once finding the derivative, set equal to zero giving you 9x^2 + 8x = 0. Factor out an x giving you x(9x + 8) = 0. You then find that your x points are 0 and -8/9. Next, you must plug in your x's to the original equation in order to find your ys. 3(0)^3 + 4(0)^2 + 5 and 3(-8/9)^3 + 4(-8/9)^2 + 5. These come out to give you the points (0,5) and (-8/9, 6.053). Box it off and that's your answer :)
Also, i understood the arc problems until we got to problems like #53 y=1/2(1/2ln x+1/x-1 + arctanx). I'm not exactly sure what you do first because there's a natural log and an arc formula. If anyone can direct me on this it'd be greatly appreciated.
I still remember the log properties from advanced math so I do okay on those problems. I also do fine on natural log problems unless they have more than one function in it. Like I said I get extremely lost then and have no idea where to start. Basically, I'm doing fine in this class except when it gets to the simplifying parts. I know it's just algebra but I just can't seem to grasp how to do all the right cancellations. I'm sure it'll take a lot of practice which I'm working on but it's just so frustrating not getting it time after time.
Finally, I want to extend the offer to arrange a study group before Wednesday to prepare for the test. I can always use a lot of help so if anyone is interested please let me know because I believe we can all help each other if we work together.
Overall, i'm looking forward to another stressful week full of work and headaches in Calculus--SIKEEEEEEE.
Week #2
Week #2 has finally come to a close and things just got harder. With the introduction of e^u and more complex derivatives, I was actually starting to struggle a little. A good example problem would be y=ln(e^x)
First off one should should identify the steps of your problem. In this case they would be
1. Natural Log
2. e^u
you problem should be 1
--- . (e^x)'
e^x
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be 1
--- . e^x . 1
e^x
After this you have to simplify algebraically, giving you e^x ,which equals 1.
----
e^x
I still mess up while simplifying because once I get a huge derivative i end up missing a sign or just plain messing up with different rules, for example the log rules. I kept thinking that log1044 cancelled leaving log101. Does anyone have any advice on remembering rules and such. It would be greatly appreciated.
First off one should should identify the steps of your problem. In this case they would be
1. Natural Log
2. e^u
you problem should be 1
--- . (e^x)'
e^x
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be 1
--- . e^x . 1
e^x
After this you have to simplify algebraically, giving you e^x ,which equals 1.
----
e^x
I still mess up while simplifying because once I get a huge derivative i end up missing a sign or just plain messing up with different rules, for example the log rules. I kept thinking that log1044 cancelled leaving log101. Does anyone have any advice on remembering rules and such. It would be greatly appreciated.
Post #2
This week in Calculus AB, things got a little harder. I do not understand how to solve a problem with many parts. I am slowly catching on, but still confused. The thing I understand most in Calculus this week was the arc trig identities.
For example, I understand how to do a problem like arccos (1/x2). First, you must know the formula for arccos, which is (1/ √1-u²) (u’). Take this problem step by step to solve it efficiently. First, find your “u,” which, in this problem, is (1/x²). Now, plug it into the first part of your equation that asks for the “u” to receive the beginning of the equation which should be (1/ (√1-(1/(x²)²). Now you must find u’, or the derivative of u. You can find this by doing the quotient rule, which is: [ (copy the bottom) (derivative of the top) – (copy the top) (derivative of the bottom) ] / (bottom)². The equation should look like ( (x²)(0) – [ (2)(1) ] ) / (x²)² = (-2x)/ (x^4) = (-2) / (x^3). You should use brackets to remind yourself when you are simplifying that you have a negative to distribute. So far, your problem should look like 1/ ( √1- (1/x^4) ) ( (-2) / (x^3) ). The rest of the problem can be solved by algebra to obtain an answer of (2/x^3√1-(1/x^4)). These types of problems are the simplest to solve for me because it only involves plugging in formulas to solve.
If anyone can help me understand order of operations for derivatives it would help me a lot. I’m not sure if you are supposed to take derivatives of everything or if you should work from inside out, then take derivative. I am just very confused on how you should work problems that include more than one rule.
For example, I understand how to do a problem like arccos (1/x2). First, you must know the formula for arccos, which is (1/ √1-u²) (u’). Take this problem step by step to solve it efficiently. First, find your “u,” which, in this problem, is (1/x²). Now, plug it into the first part of your equation that asks for the “u” to receive the beginning of the equation which should be (1/ (√1-(1/(x²)²). Now you must find u’, or the derivative of u. You can find this by doing the quotient rule, which is: [ (copy the bottom) (derivative of the top) – (copy the top) (derivative of the bottom) ] / (bottom)². The equation should look like ( (x²)(0) – [ (2)(1) ] ) / (x²)² = (-2x)/ (x^4) = (-2) / (x^3). You should use brackets to remind yourself when you are simplifying that you have a negative to distribute. So far, your problem should look like 1/ ( √1- (1/x^4) ) ( (-2) / (x^3) ). The rest of the problem can be solved by algebra to obtain an answer of (2/x^3√1-(1/x^4)). These types of problems are the simplest to solve for me because it only involves plugging in formulas to solve.
If anyone can help me understand order of operations for derivatives it would help me a lot. I’m not sure if you are supposed to take derivatives of everything or if you should work from inside out, then take derivative. I am just very confused on how you should work problems that include more than one rule.
Post #2
The second week of calculus is over! Although I think Calculus is still one of my hardest classes, it's getting better. I am understanding everything a lot better, and I've learned that it's good to ask questions.
Something I understoood this week was the natural log properties. You simply plug everthing into the formula and simplify. Let's say we have ln(x^2). You plug into the ln formula, 1/u * u', and you get 1/x^2 * 2x. When you multiply you get 2x/x^2 which simplifies to 2/x (you cancel the x's). I also understand that you may have to deal with multiple formulas in one problem. An example would be ln(x/(x^2 + 1)). In this problem you can see that you have to deal with natural log and quotient rule. You would plug it into 1/u * u' first and then into the u prime you would do the quotient rule, (vu' - uv') / v^2.
I also still get everything from last year with just normal logs. Take log 1000 = x. You would take away the log and swap 1000 and x. You then have 10^x = 1000. So from algebra you would know that 10^3 = 1000. So x = 3 would be your answer.
Something I don't understand is the difference from u^n and a^u. Take the example (ln x)^4. Would you plug it into u^n = nu^(n-1) or a^u = a^ulna * u'? I am still unsure about the whole concept.
Hope this helped somebody.
Ryan.
Something I understoood this week was the natural log properties. You simply plug everthing into the formula and simplify. Let's say we have ln(x^2). You plug into the ln formula, 1/u * u', and you get 1/x^2 * 2x. When you multiply you get 2x/x^2 which simplifies to 2/x (you cancel the x's). I also understand that you may have to deal with multiple formulas in one problem. An example would be ln(x/(x^2 + 1)). In this problem you can see that you have to deal with natural log and quotient rule. You would plug it into 1/u * u' first and then into the u prime you would do the quotient rule, (vu' - uv') / v^2.
I also still get everything from last year with just normal logs. Take log 1000 = x. You would take away the log and swap 1000 and x. You then have 10^x = 1000. So from algebra you would know that 10^3 = 1000. So x = 3 would be your answer.
Something I don't understand is the difference from u^n and a^u. Take the example (ln x)^4. Would you plug it into u^n = nu^(n-1) or a^u = a^ulna * u'? I am still unsure about the whole concept.
Hope this helped somebody.
Ryan.
Post #2
Well, to begin, week two in my opinion was much more difficult that week one. At first we reviewed the stuff from the week before, and I understood everything like the quotient and product rules. Then the next day we did the arcsin stuff, and I understood how to do that. So I'm going to show and example of it. For the problem 2arcsin(x-1), you would have to use the formula arcsinu = (1/(sqrt 1-u^2)) times u^1. To start off, you just move the 2 into the next step without doing anything, then you put the rest of the problem into the formula, but do not forget to multiply it by the derivative of u, I've done that before. So, then you get (2)(1/(sqrt1-(x-1)^2)) (1), which you further simplify to make it 2/sqrt1-(x-1)^2, then in the next step you multiply out the (x-1)^2 to get 2/(sqrt(1-(x^2)-2x+1)), then your final answer become 2/ (sqrt( -(x^2)-2x-2)). I hope you could all understand that example, I tried my best to type it, but I'm not sure if I put all the parentheses and other stuff the right way. So I understood that, and the simple log problems, like how log27 9 = 27^x = 9, then you solve for x and you find that x=2/3. I also understand how you are supposed to expand log problems like, ln2/3 = ln2-ln3. The only problems I struggle with are the extremely long problems that you have to do loads of algebra and use like 7 different rules on different sections of the problem, for example,
-[sqrt((x^2)+4)/2x^2] -1/4ln[(2+sqrt(x^2+4))/x] one thing that helped me was writing down what you had to do for each part of the problem before you solve, like how for the first part of it, you need to use quotient rule, exponent rule, and work inside exponent; in the second part, you need to take care of the ln, then use quotient rule, then exponent rule, then work the inside exponent. But I suck at doing log problems like 10log4t/t because I suck at that kind of log stuff. I think you use quotient rule, and do all kinds of simplifying after that, but I just hate it.
mher---out.
-[sqrt((x^2)+4)/2x^2] -1/4ln[(2+sqrt(x^2+4))/x] one thing that helped me was writing down what you had to do for each part of the problem before you solve, like how for the first part of it, you need to use quotient rule, exponent rule, and work inside exponent; in the second part, you need to take care of the ln, then use quotient rule, then exponent rule, then work the inside exponent. But I suck at doing log problems like 10log4t/t because I suck at that kind of log stuff. I think you use quotient rule, and do all kinds of simplifying after that, but I just hate it.
mher---out.
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