Whats derivate of x^(x^x)?

[tight sorrel]

is the derivate of x^(x^x) ..... [x^(x^x)][lnx][x^x][lnx]?

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[tight sorrel]

i got that from using the chain rule is it right?

oh wait

or is it [x^x][x^(x^x-1)][x^(x-1)][x]?

[novel flax]

Assuming you found $\dv{x}(x^x)$, you have $y=x^{x^x}\to\ln(y)=x^x\ln(x)$

[tight sorrel]

nvm i suck a twriting eq

[bright finch]

Omegabet_

[novel flax]

hence $\frac{y'}{y}=\dv{x}(x^x)\ln(x)+x^x\frac{1}{x}$

[bright finch]

Omegabet_

[novel flax]

to which the derivative follows

[tight sorrel]

and whats d/dx( x^x)

we do this again to find that?

[novel flax]

yes

[tight sorrel]

let l = x^x and take ln so ln l = x ln x

[novel flax]

yes

[tight sorrel]

ohhhhh

one more thing cant we like say x^x = u
d/dx(x^u) and use chain rule

[novel flax]

no

[tight sorrel]

why?

[novel flax]

for the same reason you cant just use chain rule on x^x

it's neither a power function nor an exponential

hence neither rule (with chain rule) works

[tight sorrel]

OH

its a fxn of x in the power of fnx of x so nothings constant as we assume in those rules?

[novel flax]

$\dv{x}x^n=nx^{n-1}$ only holds for constant $n$, and $\dv{x}a^x=a^x\ln(a)$ holds only for constant $a$

[bright finch]

Omegabet_

[novel flax]

in fact both can be 'derived' from this procedure

for example $y=x^n\to\ln(y)=n\ln(x)\to\frac{y'}{y}=\frac{n}{x}\to y'=\frac{n}{x}x^n=nx^{n-1}$

[bright finch]

Omegabet_

[tight sorrel]

OH

[novel flax]

which obviously is a massive overkill, but shows the validity in the 'simple' cases

[tight sorrel]

ohhh i get it now

thanks

+close