1. ## Calculus Derivatives, OMG!!!

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.

2. Hmmm i'm also not getting the book's answer, maybe its wrong?

3. 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.

4. Well, I did find some time to type it out, here it is:
Code:
```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

5. 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

6. 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

7. 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:
Code:
```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)]```

8. What year in school did all of you take calculus? I'm taking it now...

9. 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.

10. Originally posted by Leeman_s
What year in school did all of you take calculus? I'm taking it now...
Junior

11. 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)

12. 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!

13. 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

14. I'm taking Honors Advanced Precalc (as opposed to just normal or honors regular) as a Junior. I dunno if that counts .

15. 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...