Question

Derviative.

Answer 1

\[y = e ^{e ^{x}}\]

Answer 2

dy/dx=(e^e^x)(2e^x), is that right?

Answer 3

(2e^x) ?

Answer 4

derivative of e, is e times what ever it is raised, so u get e^e^x times e^x

Answer 5

i mean,,, (e^e^x)(e^x)

Answer 6

\[ \huge f(x)= e ^{e ^{x}} \implies f'(x)=\huge e^{e^x+x} \]

Answer 7

how u get that foolformath?, is it a rule or ?

Answer 8

apply the chain rule\[\large g(x)=e^x\]\[\large f(x)=e^{g(x)}\to f'(x)=g'(x)e^{g(x)}=e^xe^{e^x}=e^{e^x+x}\]

Answer 9

I use more elementary things, say \[ \huge y= e ^{e ^{x}} \]

Now, take logarithm (natural) of both sides,

\[ \huge \ln y= e^x \implies \frac1y \frac {dy}{dx} = e^x \]\[\huge \implies \frac {dy}{dx} = e^{e^x+x} \]

Answer 10

I (now) think Turing's process is more more elementary, as mine use implicit differentiation.

Answer 11

yeah I didn't want to be contradictory...

Answer 12

yours is faster though

Answer 13

Yes, I have a proclivity to assume that fast is elementary, but you are one with simple yet beautiful answer always :)

Answer 14

cheers :)

Answer 15

:)