Derviative.

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[y = e ^{e ^{x}}\]
dy/dx=(e^e^x)(2e^x), is that right?
(2e^x) ?

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derivative of e, is e times what ever it is raised, so u get e^e^x times e^x
i mean,,, (e^e^x)(e^x)
\[ \huge f(x)= e ^{e ^{x}} \implies f'(x)=\huge e^{e^x+x} \]
how u get that foolformath?, is it a rule or ?
apply the chain rule\[\large g(x)=e^x\]\[\large f(x)=e^{g(x)}\to f'(x)=g'(x)e^{g(x)}=e^xe^{e^x}=e^{e^x+x}\]
I use more elementary things, say \[ \huge y= e ^{e ^{x}} \] Now, take logarithm (natural) of both sides, \[ \huge \ln y= e^x \implies \frac1y \frac {dy}{dx} = e^x \]\[\huge \implies \frac {dy}{dx} = e^{e^x+x} \]
I (now) think Turing's process is more more elementary, as mine use implicit differentiation.
yeah I didn't want to be contradictory...
yours is faster though
Yes, I have a proclivity to assume that fast is elementary, but you are one with simple yet beautiful answer always :)
cheers :)
:)

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