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Math

Name: Anonymous 2008-05-29 16:45

Whats y´ of y=x^2/x+1

Name: Anonymous 2008-05-29 17:18

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

Name: Anonymous 2008-05-29 17:23

2x/x

Name: Anonymous 2008-05-30 23:39

1+1=2...amirite?

2x/1x = ?
2...amirite?

Name: Anonymous 2008-05-30 23:55

>>1
clarify your brackets bitch

Name: Anonymous 2008-05-30 23:59

>>4
>>1x
lol

Name: Anonymous 2008-05-31 13:19

guyz the derivitve is 1 use quotientz rule

Name: Anonymous 2008-05-31 16:12

x/((x+1)^2)

Just look at the fucking tables in the front/back of your math book for how to solve that easy problem, you nimwad.

Name: Anonymous 2008-05-31 16:35

>>1
y' = x/x

Name: Anonymous 2008-05-31 17:15

(dy/dx)=x

Name: Anonymous 2008-05-31 20:53

>>8
That's wrong, dumbass.

Name: Anonymous 2008-05-31 21:12

You take a function of x and you call it y
pick any x-naught that you care to try
Make a little change and call it delta-x
the corresponding change in y is what we find next

Then we take the quotient and now, carefully,
send delta-x to zero and I think you'll see
that what this gives us if our work all checks
is what we call dy/dx, it's just dy/dx!

Name: Anonymous 2008-06-01 12:12

>>12
lmao

Name: Anonymous 2008-06-03 1:00

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

Name: Anonymous 2008-06-03 5:06

It's y = x

Name: Anonymous 2008-06-05 5:10

>>12
http://www.youtube.com/watch?v=GzVSXEu0bqI&feature=related ?

Name: Anonymous 2008-06-05 15:30

y = x^2/x+1 = x^2/x + x^2/1 = x + x^2

Name: Anonymous 2008-06-06 1:42

>>16

He used to teach math classes at my Uni, sad I never got to take one from him.

Name: Anonymous 2008-06-06 18:51

>>16

Um. A/(b+c) does NOT equal A/b + A/c. That leads to bad things. :|

Name: Anonymous 2008-06-06 19:39

>>17
No.

Name: Anonymous 2008-06-06 20:57

You stupid niggers.

Name: Anonymous 2008-06-09 4:50

y = x^2/x+1
1/y = (x+1)/x^2 = x/x^2 + 1/x^2
x^2 = xy + y
??????????????????
halp

Name: Anonymous 2008-06-09 4:54

i have idea

0 = ax^2 + bx + c
x = (-b±(y^2-4ac))/2a

x^2 = xy + y
0 = -x^2 +yx + y
a = -1, b = y, c = y
x = (-y±(y^2+4y))/-2

Name: Anonymous 2008-06-09 4:55

x = (-b±(b^2-4ac))/2a

Name: Anonymous 2008-06-09 4:56

x = (-b±(b^2-4ac)^0.5)/2a
x = (-y±(y^2+4y)^0.5)/-2

oops sorry, forgot to put square root

Name: Anonymous 2008-06-11 10:06

just make it
y = X^2 . (x+1)^-1

then product rule + chain rule.

Name: Anonymous 2008-06-12 12:35

^ You can't do that

Use the quotient rule. And the force.

Name: Anonymous 2008-06-12 22:50

Sirs,

Holy shit. I'm not even in calc and I can do that. You get 1 by the quotient rule... OR THE FACT THAT X^2/X= X.

It is my humble opinion that /sci/ failed math.

(As a side note, you may want to add that the function is discontinuous at x=0, so the derivative is automatically undefined there, too)

Name: Anonymous 2008-06-16 11:45

dy/dx = (d(x^2)/dx).(d(1/(x+1))/dx)

= 2x.-1.1/(x+1)^2.1

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

Name: Anonymous 2008-06-16 17:02

Either my maths skills has failed me (unlikely):

Quotient rule:

(v(x).u'(x) - u(x).v'(x))/v^2

u (x) = x^2
u'(x) = 2x
v (x) = x+1
v'(x) = 1

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

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

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

THAT ASSUMES that the equation is (x^2)/(x+1)

BUT, if y(x) = (x^2)/(x) + 1, then the answer is 1.

Because (x^2)/(x) simplifies to x.

So y(x) = x+1

Which implies that y'(x) = 1.

Both answered and explained.

Name: Anonymous 2008-06-16 19:15

x^2 / x does not have a discontinuity at x = 0.

Name: Anonymous 2008-06-16 22:17

>>31

It is not defined at x=0.

Name: Anonymous 2008-06-16 22:24

>>26
in fact, you CAN do that. Product rule + chain rule implies quotient rule.

Name: CSharp !FFI4Mmahuk 2008-06-17 1:21

Christ, who let you people touch differentiation? You're all a bunch of retards. How hard is it to use the quotient rule? That's, like, the first fucking week of high school calculus. Jesus.

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

QhurfEdurfD.

If you want it another way, do it logarithmically (just as a proof for you relentless morons):

y=(x^2)/(x+1)
ln(y)=ln(x^2)-ln(x+1)
LOLOLchainruleandimplicitdifferentiation
y'/y=(2x)/(x^2)-(1)/(x+1)
y'/y=2/x-1/(x+1)
y'/y=(x+2)/(x^2+x)
y'=y*[(x+2)/(x^2+x)]
  =[(x^2)/(x+1)]*[(x+2)/(x^2+x)]
  =[x*(x^2+2x)]/[x*(x+1)^2]
  =(x^2+2x)/(x+1)^2

QEDOUBLE KILL!

Name: Anonymous 2008-06-17 8:42

>>32

There is no discontinuity. The value of the function x^2 / x at 0 is 0. It was said in >>28. If one function equals another, how can one have a discontinuity and the other not at the same point?

And yeah, the differentiation is trivial.

Name: Anonymous 2008-06-17 9:39

>>35
There is a discontinuity. Worthless cocknigger.

Name: Anonymous 2008-06-17 13:02

>>36


There really isn't.

If you were feeling rather pedantic you might want to try and argue there's a removable singularity there, but then it still doesn't matter because anybody who isn't a retard can see the analytic continuation.

Name: Anonymous 2008-06-17 17:45

>>35
I think I solved it for you.

x^2 / x = x

Name: Anonymous 2008-06-17 18:36

SQRT(cos^2(ϑ)-sin^2(ϑ))

Name: Anonymous 2008-06-17 19:02

>>35

That's the point, x^2/x is equal to x, except when x=0. If you take h(x)=x^2 and f(x)=x^2/x when x is not 0, and 0 elsewhere, then f and h are (Obviously) 2 diferent sets.
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