Question

find the derivative of x^x^x

Answer 1

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

Answer 2

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

Answer 3

oh god lol myin u can finish this

Answer 4

bahrom, you're taking too long, hurry up and finish it!

Answer 5

meh.. let's just use our good old wolf

Answer 6

is it?

Answer 7

no I don't think so...

Answer 8

\[y=x => y'=1 \]
\[y=x^x => \ln(y)=\ln(x^x)=> \ln(y)=x \ln (x)=> \frac{y'}{y}=\ln(x)+x \cdot \frac{1}{x}\]
\[ \text{ so } y=x^x => y'=x^x(\ln(x)+1)\]
\[ \text{ so if we have } y=x^{x^{x}} \]
\[ \text{ then } \ln(y)=\ln(x^{x^{x}})\]
\[=> \ln(y)=x^x \ln(x) \]
So we have after applying product rule while applying other rules
\[\frac{y'}{y}=x^x(\ln(x)+1) \ln(x)+x^x \cdot \frac{1}{x}\]

Answer 9

ho ho ho

Answer 10

now you solve the last equation for y' and you are done
don't forget to replace y with
\[ x^{x^{x}}\]

Answer 11

first you said y=x and then y=x^x .. i didn't get that part

Answer 12

imagine she said y and y1

Answer 13

ohh . ok

Answer 14

\[x^{x^x}=e^{x\ln(x^x)}=e^{x^2\ln(x)}\]

Answer 15

then chain rule

Answer 16

get
\[e^{x^2\ln(x)}\times (x^2\times \frac{1}{x}+2x\ln(x))\]

Answer 17

how did you transform exln(xx)=ex2ln(x) ?