How good are you at calculus?

[PolteRGeisT]

Find the derivative of this:

x^x^x

My answer so far has been

dy/dx = (x^x^x)((x^x)(lnx + 1)(lnx) + x^(x - 1))

but it's wrong.. :( I need help

[Pafnoutios]

You're right. It is wrong, but close.

d(x^x^x)/dx = x^(x^x + x - 1) * (x * ln(x)^2 + x * ln(x) + 1)

Thank you, TI-89. But I haven't been able to calculate it myself yet. y=x^x^x. Try taking the natural log of both sides, first.

[PolteRGeisT]

That's what I first tried. This is what I did.

y = x^x^x

ln(y) = ln(x^x^x)

ln(y) = (x^x)ln(x) <--- this is the iffy part for me

Note that:
y = x^x
dy/dx = (x^x)(ln(x) + 1)

Therefore:
(1/y)(dy/dx) = ((x^x)(ln(x) + 1)) * ln(x) + (x^x)(1/x) <--- product Rule

dy/dx = y * ((((x^x)(ln(x) + 1)) * ln(x)) + ((x^x)(1/x)))

However y = x^x^x

Therefore
dy/dx = (x^x^x) * ((((x^x)(ln(x) + 1)) * ln(x)) + ((x^x)(1/x)))


Using logarithmic differentiation this is what I come up with, but this appearantly wrong. I have noted the "iffy" part, perhaps that's where I made the mistake, but I can think of no other way to proceed at that step.:confused:

[PolteRGeisT]

I just graphed both of them, they are different, but very similar to each other. :(

[Pafnoutios]

OK, I've got it, now:

y = x^x^x
z = x^x
y = x^z
ln y = ln(x^z)
ln y = z ln x
(1/y)dy = z(1/x)dx + ln(x)dz
ln z = x ln x
(1/z)dz = x(1/x)dx + ln(x)dx
dz = z(1 + ln x)dx
dz = x^x(1 + ln x)dx
(1/y)dy = (x^x)(x^-1)dx + ln(x)(x^x)(1 + ln x)dx
dy = y(x^(x-1) + (x^x)(ln x)(1 + ln x))dx
dy/dx = x^(x^x)*x^(x-1)*(1 + x(ln x)(1 + ln x))
dy/dx = x^(x^x + x - 1)(x(ln x)^2 + x ln x + 1)

[Pafnoutios]

Originally posted by PolteRGeisT
My answer so far has been

dy/dx = (x^x^x)((x^x)(lnx + 1)(lnx) + x^(x - 1))



That looks like it might be the right answer. Try factoring x^(x-1) out of the binomial then distributing the (x ln x) on the inside to get what my calculator said.

[PolteRGeisT]

Thanks!!! My teacher put this on our latest calculus test but he said 0% of people got it. :eek: Thanks again :)

[PolteRGeisT]

Originally posted by Pafnoutios
That looks like it might be the right answer. Try factoring x^(x-1) out of the binomial then distributing the (x ln x) on the inside to get what my calculator said.

I graphed them all, my answer, yours, your calculators... we know that youre calculator's is correct, and your answer yielded an identicle graph. Mine is o so very close, but not quite the same. :)

[bradfordgd]

Your teacher never disclosed what the proper solution is? Do they know? Now I'm curious...

[Syngin]

Not having taken calculus since high school, I'd have to say that, after reading this thread, this is the dumbest I've felt on this forum in ages. :p

[PolteRGeisT]

Solution shall be posted Monday. :rolleyes: Why? I have no idea...

[mdwatts]

Originally posted by Syngin
Not having taken calculus since high school, I'd have to say that, after reading this thread, this is the dumbest I've felt on this forum in ages. :p

You're certainly not alone. :(

I don't think calculus was even invented back when I went to school. :p

[hard candy]

I thought calculus was on the periodic table of elements! :(

[soda_popstar]

:eek: I had the same question on one of my tests last year! I don't remember the answer though. But the posted solution seems fine. ;)

[PolteRGeisT]

Wow.... Turns out that I was right, my original equation was correct, the only reason it didn't work out in the graphing calculator was that I was putting in (x^x^x) when I should have put (x^(x^x))

Looks like the teacher was WRONG!! :D

Can't wait to show him this. hahaha

[Goran[Croatia]]

I take Geometry next year (currently in 9th grade)
:D

[DMR]

Originally posted by PolteRGeisT
Looks like the teacher was WRONG!! :D It happens- I went through the same thing once with my Boolean Logic prof in college.

If you're absolutely sure you're correct, don't let it slide, even if the professor has a problem with that. Having he/she tell who knows how many students that they're correct answer is wrong isn't helping any of you.

[DMR]

Originally posted by mdwatts
I don't think calculus was even invented back when I went to school. :p Oh, come on Mike-

Calculus was certainly around when your creators programmed you in ALGOL, wasn't it?

:D

[keyshawn]

no calc for me - yet.
[ i take Hon. ALG II\Trig this year, 11th grade; and
im taking analysis next year]

[Davy]

grrr, i just spent two hours trying to figure this out... it just kept bugging me so i HAD to try to solve it. tell us whan you found out PLEASE!!! :D

[DSwain]

I take Geometry next year (currently in 9th grade)

lol, me too. Taking Honors Geometry/Alg II

Alg I sucks btw. So simple.

[PolteRGeisT]

Ok... my teacher is a bit of an idiot... here's an alternate solution... the one he was expecting.... Note: things in [] are comments to help the reading of the math here.

y = x^(x^x))
ln (y) = ln (x^(x^x)) [take the natrual log of both sides]
ln (y) = (x^x)ln(x) [log rules make this possible]
ln ln (y) = ln ((x^x)ln(x)) [take the natural log of both sides.... again..]
((dy/dx) / (y * ln (y))) = (dx/dx) * ln (x) + x*x^(-1) + ((dx/dx) / (x * ln (x))) [take the derivative... non-calc people should just accept this part :D]
((dy/dx) / (y * ln (y))) = ln (x) + (1 / (x * ln (x))) + (1) [simplified version of the above line]
dy/dx = y * ln (y) * (ln (x) + 1 / (x * ln (x))) + 1)
dy/dx = (x^(x^x))) * ln (x^(x^x))) * (ln (x) + 1 / (x * ln (x))) + 1) [ Substitute for y]

And that is the solution. However, my shortsighted teacher thought that this was the [i]only possible method for figuring out the solution. :rolleyes:

Mr. Hughes, I shall enjoy proving you wrong :D

[PolteRGeisT]

By the way, both solutions...


dy/dx = (x^(x^x + x - 1))(x * (ln (x))^2 + x * ln (x) + 1)

dy/dx = (x^(x^x))(ln (x^(x^x)))(ln (x) + (1 / (x * ln (x))) + 1)


when graphed on a graphing calculator, yield precisely identicle values. The two graphs end up on top of each other, every value is the same. Both are correct.