Author Topic: Anyone good at Calculus (Read 1606 times)

Arnoldwanabe · « **on:** October 23, 2007, 05:06:11 AM »

Can someone tell me what the 1st and 2nd derivatives of this are: f(x) = x/(x+1)

Time is of the essence..Thanks.

ToxicAvenger · « **Reply #1 on:** October 23, 2007, 06:50:41 AM »

i'm gonna give it a shot mate but i took calc 14ish yrs ago...k here goes

bring the denominator up

(x) * -(x-1) ^ -1
(x) (-x + 1) ^ -1
(x) (x - 1)
X ^ 2 - 1
soo

dy/dx = 2x - 0 or 2x
d2y/d2x = 2 ?/

i'm prolly wrong here man...but i gave it a shot..all i remember is thatthe 1st derivative is the rate of change..and the 2nd is jerk due to gravity...

Arnoldwanabe · « **Reply #2 on:** October 23, 2007, 07:53:36 AM »

Thanks for trying bro. I'll find out today when I turn my work in.

youandme · « **Reply #3 on:** October 23, 2007, 08:06:06 AM »

do you have a TI-83 or 84? The 84 will solve derivitives, the 83 you have to work them out.

Take your time with this, cause if you can't get one step it's hard to go on to the next section.

onlyme · « **Reply #4 on:** October 23, 2007, 08:44:04 AM »

maybe I can help............ 1?

Tesla · « **Reply #5 on:** October 23, 2007, 03:26:57 PM »

Haha, going to getbig to do math homework

I'll spell it out for you.

So you've got f(x) = x/(x+1) = x*(x+1)^-1

The idea is you break f(x) into the product of two other functions, call them g(x) = x and h(x) = (x+1)^-1

f(x) = x*(x+1)^-1 = g(x) * h(x)

Then you use the product rule, and then the chain rule for h'(x)... f'(x) = g'(x)*h(x) + g(x)*h'(x) = 1*(x+1)^-1 + (-1)*x*(x+1)^-2

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

For the 2nd derivative, just use the chain rule... f''(x) = f'(f'(x)) = d((x+1)^-2)/dx

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

ToxicAvenger · « **Reply #6 on:** October 23, 2007, 04:22:29 PM »

Quote from: Tesla on October 23, 2007, 03:26:57 PM

Haha, going to getbig to do math homework I'll spell it out for you.

So you've got f(x) = x/(x+1) = x*(x+1)^-1

The idea is you break f(x) into the product of two other functions, call them g(x) = x and h(x) = (x+1)^-1

f(x) = x*(x+1)^-1 = g(x) * h(x)

Then you use the product rule, and then the chain rule for h'(x)... f'(x) = g'(x)*h(x) + g(x)*h'(x) = 1*(x+1)^-1 + (-1)*x*(x+1)^-2

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

For the 2nd derivative, just use the chain rule... f''(x) = f'(f'(x)) = d((x+1)^-2)/dx

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

oo fuck i ws wayyyy off...i hope this kid didn't turn in the mess i created...

art some point i really gotta go re take all these math classes so i dont get dumb as i age

Arnoldwanabe · « **Reply #7 on:** October 23, 2007, 04:25:14 PM »

No I didn't turn that in but I didn't turn in what Tesla did either. Oh well. Thanks anyway. And I am a little older than a "kid" just back in school to better myself. Better start sturying more though

Slin1 · « **Reply #8 on:** October 23, 2007, 04:29:51 PM »

WTF?

Is this some secret code?

Fury · « **Reply #9 on:** October 23, 2007, 04:33:16 PM »

Quote from: Slin1 on October 23, 2007, 04:29:51 PM

WTF?
Is this some secret code?

Any person not spending the majority of their life addicted to drugs should be able to understand what that is.

Anyway, learn the quotient rule Arnold. Makes it a lot easier than Tesla's way.

Slin1 · « **Reply #10 on:** October 23, 2007, 04:43:28 PM »

Quote from: BerzerkFury on October 23, 2007, 04:33:16 PM

Any person not spending the majority of their life addicted to drugs should be able to understand what that is.

Anyway, learn the quotient rule Arnold. Makes it a lot easier than Tesla's way.

Shouldn't you backed it up with the answere of the big secret code if you want to play the smart guy

btw: No addiction just enjoys the finer things in life

the choad · « **Reply #11 on:** October 23, 2007, 04:44:09 PM »

i pass calc with a D- a few years ago..Shit's as hard as Bererkfury is ugly...

ToxicAvenger · « **Reply #12 on:** October 23, 2007, 04:53:30 PM »

Quote from: Arnoldwanabe on October 23, 2007, 04:25:14 PM

No I didn't turn that in but I didn't turn in what Tesla did either. Oh well. Thanks anyway. And I am a little older than a "kid" just back in school to better myself. Better start sturying more though

yeah ya better start studying more..

find yourself a hot classmate mang!

Arnoldwanabe · « **Reply #13 on:** October 23, 2007, 05:02:01 PM »

Yeah tried that but the wife doesn't like it much when I do

Fury · « **Reply #14 on:** October 23, 2007, 05:07:25 PM »

Quote from: Slin1 on October 23, 2007, 04:43:28 PM

Shouldn't you backed it up with the answere of the big secret code if you want to play the smart guy

btw: No addiction just enjoys the finer things in life

If I recall correctly, the quotient rule would solve it as

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

Quotient rule is g(x)f'(x) - f(x)g'(x) / (g(x))^2

Do me a favor and shut the hell up now druggie.

Arnoldwanabe · « **Reply #15 on:** October 23, 2007, 05:10:39 PM »

which would then simplify to -x/(x+1)...that's what I got!

MB_722 · « **Reply #16 on:** October 23, 2007, 05:59:44 PM »

I remember this stuff.

youandme · « **Reply #17 on:** October 23, 2007, 06:09:48 PM »

Leave it to TESLA

haider · « **Reply #18 on:** October 23, 2007, 06:20:39 PM »

Noobs.

Slin1 · « **Reply #19 on:** October 24, 2007, 05:00:51 PM »

Quote from: haider on October 23, 2007, 06:20:39 PM

Noobs.

Haha yes, live this noobs with there guessing

Arnoldwanabe · « **Reply #20 on:** October 25, 2007, 06:11:37 AM »

Quote from: Slin1 on October 24, 2007, 05:00:51 PM

Haha yes, live this noobs with there guessing

WTF...is THIS some secret code?