Question

Not really a question.
Just thinking about a calculus concept.
Comment if you like. :)

Answer 1

Here are some common derivative rules that we know.
For these I'm assuming f=f(x), and a is constant.
\[\Large\rm y=f^a,\qquad y'=a(f^{a-1})f'\qquad\qquad \text{Power Rule}\]\[\Large\rm y=a^f,\qquad y'=a^f (f') \ln a\qquad\qquad \text{Exponential Rule}\]

I'm just trying to think about what happens when we have a function to a function and take it's derivative and then see if I can come up with some generalization.
You guys can add insight if you want, I just want to show my thought process.
So for some functions f=f(x), g=g(x),\[\Large\rm y=f^g\]log of each side,\[\Large\rm \ln y=g \ln f\]Taking derivative,\[\Large\rm \frac{1}{y}y'=g' \ln f+g \frac{1}{f}f'\]Multiply y to the other side,\[\Large\rm y'=y\left[g' \ln f+g \frac{1}{f}f'\right]\]plug in the y, get a common denominator in the brackets,\[\Large\rm y'=f^g\left[\frac{f g' \ln f+f' g}{f}\right]\]Let's ummm, pull the 1/f out and combine it with the term in front,\[\Large\rm y'=f^{g-1}\left[f g' \ln f+f' g\right]\]Then let's distribute the thing into the brackets,\[\Large\rm y'=f^g(g')\ln f+g(f^{g-1})f'\]Now that I'm looking at it, I can see that getting the common denominator and pulling out the 1/f was maybe a couple of unnecessary steps hehe.


Anyway, check this out, this is kind of neat.
\[\Large\rm y'=\text{Exponential Rule + Power Rule}\]We started out using product rule and a bunch of fun stuff, and it looks like we end up with the sum of these two rules that we listed at the start.
Hmm interestinggggg.

Answer 2

Maybe this is already detailed on wikipedia or somewhere, I dunno.
I just wanted to work through it and try to make some kind of connection :p

Answer 3

I think it works minus the y=exponential rule +power rule part

Answer 4

No? :o
It looks like it matches those rules.. hmm

Answer 5

it's a combination of em, not a straight addition of the two

Answer 6

nope i retract. I was reading product rule for some reason

Answer 7

So let's try using that on a simple example problem maybe,
\[\Large\rm y=x^{2x}\]without using log of each side and all that business.\[\Large\rm y=\text{Exponential Rule + Power Rule}\]\[\Large\rm y'=x^{2x}(2x)'\ln x+2x(x^{2x-1})x'\]\[\Large\rm y'=2x^{2x}\ln x+2x(x^{2x-1})\]\[\Large\rm y'=2x^{2x}\left[\ln x+1\right]\]Which is the correct answer.
Hmm that's kinda neat :)

Answer 8

yea, nice derivation

Answer 9

Maybe I'm being a bit sloppy calling it "exponential rule" when our base is clearly x, but we're treating it as constant for that step, i dunno

Answer 10

I assume there are some limits to the function though

Answer 11

..... So much math *dies* I can't take it.... all the big words... ahhhhhh

Answer 12

XD

Answer 13

might be a only works for positive f type of thing... I'm not sure

Answer 14

ya probably, something like this ends up being crazy anyway,\[\Large\rm (-2)^x\]even without taking its derivative hehe

Answer 15

shall we find out?

Answer 16

:x

Answer 17

Oh wolfram, how I love thee http://www.wolframalpha.com/input/?i=%28%28-2%29%5Ex%29%27

Answer 18

so, have to dip into complex analysis to pull that bad boy off

Answer 19

Ohh interesting

Answer 20

\[\Large\rm y=(-2)^x\]\[\Large\rm y'=(-2)^x \ln(-2)\qquad\qquad \text{Exponential Rule}\]\[\Large\rm y'=(-2)^x\left[\ln(2)+\ln(-1)\right]\]And yah gotta dip into the complex here.. hmm trying to remember >.<

Answer 21

i pi

Answer 22

Oh oh oh right!\[\Large\rm e^{i \pi}=-1\]\[\Large\rm \ln e^{i \pi}=\ln(-1)\]\[\Large\rm i \pi=\ln(-1)\]

Answer 23

Hmm neato :)

Answer 24

yea, cool trick

Answer 25

wanna check my theory? I may need to prove it and get published if it works

Answer 26

maybeeee <.< I'm feeling kinda lazy right now lol

Answer 27

haha, well, I tagged ya. It's all about coloring :D

Answer 28

Interesting, I have always noticed that there was this thing \[\large \frac{d}{dx} x^x = xx^{x-1}+x^x \ln x\] but I never really thought much past that, i was to distracted while taking calculus 2 haha.

Answer 29

Ok let's see if we can take this a step further, what about this? \[\Large y=f^{g^h}\]

Answer 30

buhaha >.> of course YOU would do something like that lol

Answer 31

... well, question cause I'm not sure, isn't that the same as \(lny=hglnf\)?

Answer 32

so it's just a triple iteration of the product rule?

Answer 33

Nah it's more like \[\ln( \ln y)=h \ln g + \ln(\ln f)\]

Answer 34

\[\Large\rm \ln y=g^h \ln f\]If we're allowed to use our first result, then I don't think we need to log again.

Answer 35

Oh true

Answer 36

But if you want to start from scratch, yah it's a bit of work :)

Answer 37

so that's where I'm wondering cause we'd have \[lny=ln f^{g^h}\]

Answer 38

so we can bring the exponents out front right?

Answer 39

but can we pull both...I actually don't know

Answer 40

Fib, I think you were confusing \(\Large\rm \left(f^g\right)^h\) with \(\Large\rm f^{(g^h)}\).
We want the second one :D

Answer 41

oh

Answer 42

yep

Answer 43

So I guess this is getting gross but applied twice: \[\large y' = (g^h)f^{g^h-1}f'+(g^h)'f^{g^h} \\ \large y' = (g^h)f^{g^h-1}f'+(hg^{h-1}g'+g^hh')f^{g^h}\]

Answer 44

I agree with kainui now

Answer 45

So I guess this is getting gross but applied twice: \[\large y =f ^{g^h} \\ \large y' = (g^h)f^{g^h-1}f'+(g^h)'f^{g^h} \\ \large y' = (g^h)f^{g^h-1}f'+(hg^{h-1}g'+g^hh')f^{g^h}\]

Answer 46

it seems recursive

Answer 47

Ooo interesting :3

Answer 48

Uh oh the site is tweaking out on me :(
Too much LaTeX in this thread maybe lol

Answer 49

lol potentially

Answer 50

Yeah so we could keep doing this for \[\large y=f^{g^{h^\cdots}}\] So now what if we did it infinitely, what would the series be of terms in the derivative?

Answer 51

...That would be a serious series...

Answer 52

did u figure out the pattern for nth exponent functions of functions

Answer 53

Maybe we should index them \[\Large y=f_1^{f_2^{f_3^\cdots}}\]

Answer 54

ya lol that looks a lot neater

Answer 55

the recursion is always coming from the exponential rule side, i think its better to just see the pattern with lns but i guess that wud defeat the purpose

Answer 56

we need new ntoation for mulitple exponents