Is it possible for a function, say f(x), to have a derivative

>>8858469
wew

>>8858469
f(x) = 0

>>8858469
e^x

>>8858482

this

idk man u might be on to something
try (2.7182818284590452353602874713527)^x and see what that gets you

Try 2.718281828459045235360287471352662497757247093699959574966967627724076630353547594571382178525166427427466391932003059921817413596629043572900334295260595630738132328627943490763233829880753195251019011573834187930702154089149934884167509244761460668082264800168477411853742345442437107539077744992069551702761838606261331384583000752044933826560297606737113200709328709127443747047230696977209310141692836819025515108657463772111252389784425056953696770785449969967946864454905987931636889230098793127736178215424999229576351482208269895193668033182528869398496465105820939239829488793320362509443117301238197068416140397019837679320683282376464804295311802328782509819455815301756717361332069811250996181881593041690351598888519345807273866738589422879228499892086805825749279610484198444363463244968487560233624827041978623209002160990235304369941849146314093431738143640546253152096183690888707016768396424378140592714563549061303107208510383750510115747704171898610687396965521267154688957035035402123407849819334321068170121005627880235193033224745015853904730419957777093503660416997329725088687696640355570716226844716256079882651787134195124665201030592123667719432527867539855894489697096409754591856956380236370162112047742722836489613422516445078182442352948636372141740238893441247963574370263755294448337998016125492278509257782562092622648326277933386566481627725164019105900491644998289315056604725802778631864155195653244258698294695930801915298721172556347546396447910145904090586298496791287406870504895858671747985466775757320568128845920541334053922000113786300945560688166740016984205580403363795376452030402432256613527836951177883863874439662532249850654995886234281899707733276171783928034946501434558897071942586398772754710962953741521115136835062752602326484728703920764310059584116612054529703023647254929666938115137322753645098889031360205724817658511806303644281231496550704751025446501172721155519486685080036853228183152196003735625279449515828418829478761085^x

>>8858482
don't forget c*e^x

>>8858469
f(x) = 1 + x + (x^2)/2 + (x^3)/6 + (x^4)/24 + x^5/120 + ... + (x^n)/(n!) + ...

>>8858469

A more interesting question is to find a function such that any fractional derivaite is equal to the original function.

f(x) = f'(x)

really not hard

>>8858664
Kek

Without looking it up, find a function such that f'(x) = f^-1 (x).

>>8859155
(2/a)sqrt(ax) for all a in the reals.

>>8858469
y=(dy/dx)
dy/y=dx
integrate both sides
ln|y|=x+c
y=ce^x

>>8859164
Huh? No, that's not correct.

>>8858469
try again leonhard

>>8859164
So taking a to be 1,

>f'(x) = 1/sqrt(x)
>f^-1(x) = (x^2)/4
No.

The expression f^-1(x) is the inverse function of f(x) such that f^-1(f(x)) = x. Like arcsine is to sine or like nth power is to nth root.

>>8859155
[math]f(x) = \varphi^{-\frac{1}{\varphi}} x^\varphi}[/math]

>>8859234
Shit.

f(x) = \varphi^{-\frac{1}{\varphi}} x^{\varphi}}

i think it's cool that you explored euler's number in a roundabout way and got to 2.7

>>8859241
If this doesn't work I'm giving up.

[math]f(x) = \varphi^{-\frac{1}{\varphi}} x^{\varphi}}[/math]

if u take the derivative of sin(x) 4 times you get sin(x). Does that count?

>>8858469
nice b8 m8

>>8859155
f=c*x^n
f'=c*n*x^(n-1)
f^(-1)=1/c*x^(1/n)
c*n*x^(n-1)=1/c*x^(1/n)
c^2*n*x^(n-1-1/n)=1
Only possible if coefficient is equal to 1 and exponent is equal to 0
c^2*n=1
c^2=1/n
n-1-1/n=0
n^2-n-1=0
n=(1+-sqrt(5))/2
c=n^(-.5)

>>8858498
oh wow it worked

>>8859245
[math] f(x)=\varphi^{-\frac{1}{\varphi}}x^{\varphi}} [/math]

>>8860944
[math]f(x) = \phi^{-\frac{1}{\phi}} x^{\phi}
[/math]

>>8858469
whoa, slow down euler, i think you might be onto something