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View Full Version : Calculus Derivatives, OMG!!!

Xei
10-23-2003, 02:14 AM
That's it! I'm so damn frusterated I... I.. AHHH!!

So, it's 2:07am and I work in afew hours. I just finished my derivatives exam through online school and I got 100%. But that's not the point, the point is that there are still some questions that are ........ing me off. For instance:

The book says:

"Find the second derivative of: ((x^2)+1)^(1/2)"

Okay, "Simple", I think. I found the first derivative to be:

x((x^2)+1)^(-1/2)

The book agrees with that answer, except for the second derivative. The book believes that the second derivative is ((x^2)+1)^(-3/2) which makes no sense at all. How is this possible? I tried using the Product Rule w/ Chain Rule which would normally go like this:

X = V
((x^2)+1)^(-1/2) = U

V ' = Derivative of V
U ' = Derivative of U

So now, my derivative should be:

(V ')(U) + (U ')(V) = 0

Which is quite long for me to type out, but anyways, I end up with an answer which should include X^2, but it doesn't. I'm frusterated, this makes no sense. Please help.

Clyde
10-23-2003, 04:38 AM
Hmmm i'm also not getting the book's answer, maybe its wrong?

axon
10-23-2003, 06:26 AM
The book is correct my friend...I'm really short of time right now, so I will post the proper way to derive f''(x) when I get back in front of the computer.

But I have done the problem and the book is indeed right.

axon
10-23-2003, 06:53 AM
Well, I did find some time to type it out, here it is:

we agree that

f'(x) = x / (x^2+1)^(1/2)

Now, to compute f''(x) you can bring the denominator up or leave it there.
If you leave it, the derivation will be a bit harder, but simplification
much easier.

f''(x) =[ (x^2+1)^(1/2) - x(x^2+1)^(-1/2) ] / (x^2 + 1)
//to simplify brek into two fractions
=[(x^2+1)^(1/2) / (x^2+1) ] - [ x / {(x^2+1)^(1/2)(x^2+1)}
//from now on its just algebra, take common denoms
= (x^2 + 1 - x^2) / [(x^2+1)^(1/2)(x^2+1)]
//x^2 in the numerator cancel out and the denom simplifies to...
= 1 / (x^2+1)^(3/2)

often times the hard part of calculus is the agebra...as weird as it sounds. I know that my professors on exams accepted not simplified solutions. If you have a TI calculator, you could put in your solution and equate it with the book's, and see if you get a 'true'. That way you'll know if you have the same answer just unsimplified.

axon

EDIT:: write what i 'coded' on paper; it will make more sense

Clyde
10-23-2003, 07:37 AM
Ah i see, i got it to be -x^2(x^2+1)^-3/2 + (x^2 + 1)^-1/2 which is unsimplified like you said.

Normally when i have to use calculus it's to calculate numerical answers so i don't have to simplify stuff. I better practise my algebra evidently its getting rather rusty :o

whistlenm1
10-23-2003, 09:54 AM
f(x) = (x^2 + 1)^(1/2)

f'(x) = x(x^2 + 1)^(-1/2)

f''(x) = [(x^2 + 1)^(1/2) - x^2(X^2 + 1)^(-1/2)]/[((x^2 + 1)^(1/2))^2]

= (x^2 + 1)^(-1/2) * [((x^2 + 1) - X^2)/(x^2 + 1)

= (x^2 + 1)^(-1/2) * [1/(x^2 + 1)]

or

= 1/(x^2 +1)^(1/2) *1/(X^2 + 1)^(1)

= 1/(x^2 + 1)^(3/2) or (x^2 + 1)^(-3/2)

I believe :p

UnregdRegd
10-23-2003, 04:22 PM
Math is tricky and requires an attentive eye. I was wondering why I had come up with a different solution until I noticed I had missed a second x:

f''(x) = [(x^2 +1)^(-1/2)] -(1/2)x[(x^2 +1)^(-3/2)](2x)

f''(x) != 1/[(x^2 +1)^(1/2)] -x/[(x^2 +1)(x^2 +1)^(1/2)]
f''(x) != (x^2 -x +1)/[(x^2 +1)(x^2 +1)^(1/2)]

f''(x) = 1/[(x^2 +1)(x^2 +1)^(1/2)]

Leeman_s
10-23-2003, 08:04 PM
What year in school did all of you take calculus? I'm taking it now...

axon
10-23-2003, 08:13 PM
Originally posted by Leeman_s
What year in school did all of you take calculus? I'm taking it now...

I took it first semester university.

golfinguy4
10-23-2003, 09:40 PM
Originally posted by Leeman_s
What year in school did all of you take calculus? I'm taking it now...

Junior

confuted
10-23-2003, 10:09 PM
Originally posted by Leeman_s
What year in school did all of you take calculus? I'm taking it now...

I took it junior year, and I'm taking it again senior year (AP Calc BC now... before it was just the first 7 chapters of the calc book)

whistlenm1
10-24-2003, 06:39 AM
a little over a year, and I have been tutoring eversince off and on again. The nightmares are gone but now I'm left with calc dreams!

Leeman_s
10-24-2003, 08:44 AM
Wow, at my HS only four people are taking calc right now that are juniors. We must have on average stupid people here where I live

Speedy5
10-24-2003, 01:21 PM
I'm taking Honors Advanced Precalc (as opposed to just normal or honors regular) as a Junior. I dunno if that counts :).

axon
10-24-2003, 01:24 PM
Originally posted by Speedy5
I'm taking Honors Advanced Precalc (as opposed to just normal or honors regular) as a Junior. I dunno if that counts :).

sorry, it doesn't...:) thats a course similar to what I took as highest in high school...now that I look back on it, it was nothing more than a waste of my time...

Zach L.
10-24-2003, 09:43 PM
Took AP Calc BC sophomore year, and then Calc III the following semester. Now I get to enjoy the "simplicity" of abstract algebra (really cool stuff).

frenchfry164
10-25-2003, 12:37 PM
Wow, at my HS only four people are taking calc right now that are juniors. We must have on average stupid people here where I live
My HS doesn't even offer pre-calc until your senior year, and if you want to take calculus 1 in HS, you have to sacrifice an elective to take it.

10-26-2003, 12:14 PM
hmmm.... well, i am not worried about my local Highschools having calculus or not, because being in college at age 16, i am taking Calc I and II this year.

Now, alittle off topic, but since im not going to the best school I could, for financial reasons(arkansas technical university, one of the better schools for comp sci in the region atleast.), im planning on transfering to cal-tech or MIT(better pray for scholarship as its 40K a year) as a freshman after my sophomore year here is finished.

golfinguy4
10-26-2003, 06:22 PM
No offense, unless you are some ubergenius, I wouldn't plan on transering to Caltech or MIT. Those schools are very hard to get into and they reject multiple 1600s every year.

However, if you do get into MIT, you won't have to worry about the financial aspect. At MIT, they calculuate how much you can afford to pay...then, they pay the rest in grants.