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anonymous

  • 4 years ago

Derviative.

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  1. anonymous
    • 4 years ago
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    \[y = e ^{e ^{x}}\]

  2. anonymous
    • 4 years ago
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    dy/dx=(e^e^x)(2e^x), is that right?

  3. anonymous
    • 4 years ago
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    (2e^x) ?

  4. anonymous
    • 4 years ago
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    derivative of e, is e times what ever it is raised, so u get e^e^x times e^x

  5. anonymous
    • 4 years ago
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    i mean,,, (e^e^x)(e^x)

  6. anonymous
    • 4 years ago
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    \[ \huge f(x)= e ^{e ^{x}} \implies f'(x)=\huge e^{e^x+x} \]

  7. anonymous
    • 4 years ago
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    how u get that foolformath?, is it a rule or ?

  8. TuringTest
    • 4 years ago
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    apply the chain rule\[\large g(x)=e^x\]\[\large f(x)=e^{g(x)}\to f'(x)=g'(x)e^{g(x)}=e^xe^{e^x}=e^{e^x+x}\]

  9. anonymous
    • 4 years ago
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    I use more elementary things, say \[ \huge y= e ^{e ^{x}} \] Now, take logarithm (natural) of both sides, \[ \huge \ln y= e^x \implies \frac1y \frac {dy}{dx} = e^x \]\[\huge \implies \frac {dy}{dx} = e^{e^x+x} \]

  10. anonymous
    • 4 years ago
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    I (now) think Turing's process is more more elementary, as mine use implicit differentiation.

  11. TuringTest
    • 4 years ago
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    yeah I didn't want to be contradictory...

  12. TuringTest
    • 4 years ago
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    yours is faster though

  13. anonymous
    • 4 years ago
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    Yes, I have a proclivity to assume that fast is elementary, but you are one with simple yet beautiful answer always :)

  14. TuringTest
    • 4 years ago
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    cheers :)

  15. anonymous
    • 4 years ago
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    :)

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