Calculus Derivatives, OMG!!!

[Xei]

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]

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

[axon]

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]

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]

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

[whistlenm1]

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

[UnregdRegd]

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]

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

[axon]

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]

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

Junior

[confuted]

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]

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]

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]

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

[axon]

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...