Grr @ Calculus

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macdude425

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Okay, so I've got this take home test to do by Monday morning. I've got two problems left, and I've hit a brick wall.

Here's what the problems are.

1) Find the first and second derivative, and critical points, of y=(x-1)/(x^2+1)
(Note: the second derivative critical numbers will be discovered using synthetic or long division)

Already got the first derivative on this one, but I keep getting erratic answers on the second one when I try to simplify it.

2) Find the first and second derivatives and critical numbers of y= cos(x) + sin(2x) on [0,2pi]
(hint: to find the critical numbers of the first derivative you will need to use a double angle formula to force a quadratic "type" equation - use the substitution u=sin(x), remember the interval is from [0,2pi] so you might want to consider all the solutions in degrees and then convert to radians)

I already have the first and second derivatives on that second one; -sin(x) + 2cos(2x) and -cos(x) - 4sin(2x) respectively, but I'm not exactly sure WHAT to do from there to get the critical numbers from the first deriv.

Any suggestions?

Thanks.
 
d2y/dx2[f/g]=(g*f'-f*g')/g^2: plug-n-chug Are you a junior or senior? I learned this just last week.
for critical numbers, i forgot this, my teacher gave it one moments mention and went on: is it dy/dx=0, solve? or isn't it the max and mins?
 
Okay, so I've got this take home test to do by Monday morning. I've got two problems left, and I've hit a brick wall.

Here's what the problems are.

1) Find the first and second derivative, and critical points, of y=(x-1)/(x^2+1)
(Note: the second derivative critical numbers will be discovered using synthetic or long division)

Already got the first derivative on this one, but I keep getting erratic answers on the second one when I try to simplify it.

2) Find the first and second derivatives and critical numbers of y= cos(x) + sin(2x) on [0,2pi]
(hint: to find the critical numbers of the first derivative you will need to use a double angle formula to force a quadratic "type" equation - use the substitution u=sin(x), remember the interval is from [0,2pi] so you might want to consider all the solutions in degrees and then convert to radians)

I already have the first and second derivatives on that second one; -sin(x) + 2cos(2x) and -cos(x) - 4sin(2x) respectively, but I'm not exactly sure WHAT to do from there to get the critical numbers from the first deriv.

Any suggestions?

Thanks.
Well Im in Calc one as well. But Im not doing so great, mostly because of the applications of derivatives and applied maximum and minmum sections, BUT I do have the derivatives and first/second derivatives test down.
Anyways, I can try to assist..


1. I have no idea why the instructions tell you to use synthetic devision to find critcal numbers of first/derivative.. Further more, I was not taught that way. Simply get the first and second derivatives and factor them to find critcal points.. If they are not easily factorable you can use Netwons method and a Calculator graph to assist...

2. I am horrible with trig functions and finding critical points for them, however, you will have to reference alot of pre calc for this problem in which you find the entries to the functions that will equal zero, which will be alot of odd/even multiples of pie en such... Depending on the specific trig function of course, furthermore I never had any such problem for this section where I had to use double angle forumulas to solve... But, I think, if you have the properties of the trig functions in front of you, you could figure it out.. So from the point that you already have the first and second derivatives for such function you can look back at what entries (weather it be odd/even multiples of Pie, or something else) that make given function == to zero...

I hope that helps somewhat...
 
d2y/dx2[f/g]=(g*f'-f*g')/g^2: plug-n-chug Are you a junior or senior? I learned this just last week.
for critical numbers, i forgot this, my teacher gave it one moments mention and went on: is it dy/dx=0, solve? or isn't it the max and mins?
I'm a senior - and you got that right, in both cases, but my problem is I can't get it to simplify consistently - I see three different ways to simplify it but none of them provide the same answer as the non-simplified derivative - I'll post what I have tomorrow.
 
My teacher is at least 5 sections ahead of all the other calc classes at my school. It sucks. I wish I was back on your section.. Your going to want to end your life when you get to applications of derivatives...
 
My teacher is at least 5 sections ahead of all the other calc classes at my school. It sucks. I wish I was back on your section.. Your going to want to end your life when you get to applications of derivatives...

My teacher is starting next week...
 
My teacher is at least 5 sections ahead of all the other calc classes at my school. It sucks. I wish I was back on your section.. Your going to want to end your life when you get to applications of derivatives...
LOL, so noted.

Okay, here's what I got for the first derivative, simplified.
y' = (-x^2+2x+1)/(x^2+1)^2

Now, for the second, I got this unsimplified....
y'' = [(x^2+1)^2*(-2x+2) - (x^2+2x+1)*(2(x^2+1)*2x)]/(x^2+1)^4

If I simplify, I get this...
(-2x^3+6x^2-14x+2)/(X^2+1)^3

...and the two do not match. If we let x=2, you get totally different answers on the simplified and non simplified answers.

Any more help? Thanks (I'm starting to get desperate).
 
I do not understand why have this urge to combine terms. Its Far easier to find zeros of a function when everything is unfolied..

Anyways, in I am going to do the problem from the start and post back to see if I can help you, so give me a few mins..
 
Oh wow, editing.. I sure as heck hope noone viewed my first attempt at that problem...

Anyways,
F' == ((x^2+1)-2x(x-1)/(X^2+1)^2
F' ===((x^2+1)-(2x^2-2x))/(x^4+2x^2+1)
F' ==== (-x^2-2x+1)/(x^4+2x^2+1)

So the numerator does not factor nicely at all. So Using a graphing calculator to approx the zeros. They are 0.41421356 and -2.414214

So now Onto F'' .
F'' == (2x^5+8x^4-4x^3-5x^2-1)/((x^4+2x^2+1)^2)
And again the numerator does not factor easily so I found zeros to be.
-4.327251, -0.7570148 and 1.

Now, If your first and second derivatives are correct, which Im not sure if they are. Then -1 is the zero of F' and
0.1522924 is the zero of F''. But if your confident in the derivatives youve found then you can simply use a calculator to find approx zeros..
 
mmm
f' = 0 gives stationary points... -x^2+2x+1=0 gives √8 and -√8
let me work on f''

unless i completly got is wrong..which is possible
 
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