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AntiDerrivative of X^X?

Name: Anonymous 2007-04-04 21:20 ID:V5c0tLJl

That is X to the power of X.

My friend was actually asked this in a job interview..  not just for the answer but also how to find it.

Name: Anonymous 2007-04-04 21:28 ID:Heaven

(x^(x+1))/(x+1) - raise the exponent by one and divide by the new value.

Perhaps his potential employer wanted an applicant with basic high school calc in his repertoire?

Name: Anonymous 2007-04-04 21:48 ID:HEkzJOjz

>>2
wrong, that rule only works for constant exponents.

Name: Anonymous 2007-04-04 21:51 ID:HEkzJOjz

>>1
what job was he applying for?

Name: Anonymous 2007-04-04 21:57 ID:zGlb1ogK

hmmm...

(1/(ln x)) (x^x)?

Name: Anonymous 2007-04-04 22:02 ID:zGlb1ogK

This is >>5

never mind I don't think that would work.  I'm thinking x^x isn't solvable analytically.

Name: Anonymous 2007-04-04 22:03 ID:Heaven

>>2
epic fail

Anyways it's ln(x)*x^x + x^x. You can find this out by implicitly differentiating ln(y) = x*ln(x).

Name: Anonymous 2007-04-04 22:03 ID:Heaven

>>6
Ahahaha oh god. Is this high school drop out night or something?

Name: Anonymous 2007-04-04 22:08 ID:HEkzJOjz

>>7
wrong, try differentiating ln(x)*x^x + x^x, i assure you it is not x^x.

using mathematica's integrator, it says there probably is not an explicit formula.

I dunno, maybe find a power series expansion?

Name: Anonymous 2007-04-05 0:46 ID:nihlWjky

Try this site that uses Mathematica:
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic

Name: Anonymous 2007-04-05 5:49 ID:3ENZoKfn

>>6
*Ding ding ding*, we have winner.

Rule of thumb: All questions in job interviews are trick questions.

Name: Anonymous 2007-04-05 7:07 ID:CBf4EofY

But there isn't a tried and true method to check if a given function has an elementary anti-derivative, right?

Name: Anonymous 2007-04-05 8:16 ID:BLy/Vom/

the answer to your problem:
moredhel.za.pl/pics/losers/images/0095_tikov4_jpg.jpg

Name: Anonymous 2007-04-05 16:51 ID:5VGFe1Bl

>>13
To respond to you in a method that your people are accustomed to "WHAT-eveeeeer".

Name: Anonymous 2007-04-05 17:37 ID:WoRmssDA

>>9
got it right.

Name: Anonymous 2008-06-19 4:17

This is an in-line \begin{math}\int\frac
{d\theta}{1+\theta^2} = \tan^{-1}
\theta+ C\end{math} equation.

Name: Anonymous 2008-06-19 4:18

This is an in-line \int\frac
{d\theta}{1+\theta^2} = \tan^{-1}
\theta+ C equation.

Name: Anonymous 2008-06-19 4:20

x_+ = \frac{-b + \sqrt {b^2-4ac}}{2a}</math>
| style="width:100px" align="center" | and
|<math>\ x_- = \frac{-b - \sqrt {b^2-4ac}}{2a}

Name: Anonymous 2008-06-19 4:20

x_+ = \frac{-b + \sqrt {b^2-4ac}}{2a} \ x_- = \frac{-b - \sqrt {b^2-4ac}}{2a}
Don't change these.
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