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Getbig Main Boards => General Topics => Topic started by: Arnoldwanabe on October 23, 2007, 05:06:11 AM

Title: Anyone good at Calculus
Post by: 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.
Title: Re: Anyone good at Calculus
Post by: ToxicAvenger 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...  :(
Title: Re: Anyone good at Calculus
Post by: Arnoldwanabe on October 23, 2007, 07:53:36 AM
Thanks for trying bro.  I'll find out today when I turn my work in.
Title: Re: Anyone good at Calculus
Post by: youandme 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.
Title: Re: Anyone good at Calculus
Post by: onlyme on October 23, 2007, 08:44:04 AM
maybe I can help............ 1?
Title: Re: Anyone good at Calculus
Post by: Tesla on October 23, 2007, 03:26:57 PM
Haha, going to getbig to do math homework  8)   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
Title: Re: Anyone good at Calculus
Post by: ToxicAvenger on October 23, 2007, 04:22:29 PM
Haha, going to getbig to do math homework  8)   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 :(
Title: Re: Anyone good at Calculus
Post by: 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  ???
Title: Re: Anyone good at Calculus
Post by: Slin1 on October 23, 2007, 04:29:51 PM
WTF?



Is this some secret code?
Title: Re: Anyone good at Calculus
Post by: Fury on October 23, 2007, 04:33:16 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.
Title: Re: Anyone good at Calculus
Post by: Slin1 on October 23, 2007, 04:43:28 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   :o


btw: No addiction just enjoys the finer things in life
Title: Re: Anyone good at Calculus
Post by: the choad 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... :o
Title: Re: Anyone good at Calculus
Post by: ToxicAvenger on October 23, 2007, 04:53:30 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!
Title: Re: Anyone good at Calculus
Post by: Arnoldwanabe on October 23, 2007, 05:02:01 PM
Yeah tried that but the wife doesn't like it much when I do  ;D
Title: Re: Anyone good at Calculus
Post by: Fury on October 23, 2007, 05:07:25 PM
Shouldn't you backed it up with the answere of the big secret code if you want to play the smart guy   :o


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.  ;D ;D
Title: Re: Anyone good at Calculus
Post by: Arnoldwanabe on October 23, 2007, 05:10:39 PM
which would then simplify to -x/(x+1)...that's what I got!
Title: Re: Anyone good at Calculus
Post by: MB_722 on October 23, 2007, 05:59:44 PM
I remember this stuff.
Title: Re: Anyone good at Calculus
Post by: youandme on October 23, 2007, 06:09:48 PM
Leave it to TESLA

 ;D
Title: Re: Anyone good at Calculus
Post by: haider on October 23, 2007, 06:20:39 PM
Noobs.
Title: Re: Anyone good at Calculus
Post by: Slin1 on October 24, 2007, 05:00:51 PM
Noobs.

Haha yes, live this noobs with there guessing
Title: Re: Anyone good at Calculus
Post by: Arnoldwanabe on October 25, 2007, 06:11:37 AM
Haha yes, live this noobs with there guessing

WTF...is THIS some secret code?