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anonymous
 4 years ago
Derviative.
anonymous
 4 years ago
Derviative.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dy/dx=(e^e^x)(2e^x), is that right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0derivative of e, is e times what ever it is raised, so u get e^e^x times e^x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i mean,,, (e^e^x)(e^x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \huge f(x)= e ^{e ^{x}} \implies f'(x)=\huge e^{e^x+x} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how u get that foolformath?, is it a rule or ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1apply the chain rule\[\large g(x)=e^x\]\[\large f(x)=e^{g(x)}\to f'(x)=g'(x)e^{g(x)}=e^xe^{e^x}=e^{e^x+x}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I use more elementary things, say \[ \huge y= e ^{e ^{x}} \] Now, take logarithm (natural) of both sides, \[ \huge \ln y= e^x \implies \frac1y \frac {dy}{dx} = e^x \]\[\huge \implies \frac {dy}{dx} = e^{e^x+x} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I (now) think Turing's process is more more elementary, as mine use implicit differentiation.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah I didn't want to be contradictory...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yours is faster though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, I have a proclivity to assume that fast is elementary, but you are one with simple yet beautiful answer always :)
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