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PhoenixFire

  • 2 years ago

How do you find the derivative of \[{cos(x)}^{cos(x)}\]

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  1. lgbasallote
    • 2 years ago
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    interesting

  2. lgbasallote
    • 2 years ago
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    is x raised to cos x? or is cos x raised to cos x?

  3. PhoenixFire
    • 2 years ago
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    cos(x) raised to cos(x). \[(cos(x))^{cos(x)}\]

  4. Chlorophyll
    • 2 years ago
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    Let y = ( cosx )^( cosx) Take ln both sides:...

  5. PhoenixFire
    • 2 years ago
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    \[y=(cos x)^{cos x}\] \[ln y=(cos x)(ln(cos x))\] differentiate both side with respect to x. \[LHS={1 \over y}{dy \over dx}\] RHS=product rule. then rearrange for dy/dx? Yeah?

  6. Chlorophyll
    • 2 years ago
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    Yupe :)

  7. Chlorophyll
    • 2 years ago
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    Actually, ( lny )' = y' / y

  8. PhoenixFire
    • 2 years ago
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    \[{dy \over dx}=-ysin x(ln(cos x) + 1)\] That doesn't seem right.

  9. Chlorophyll
    • 2 years ago
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    Then plug y = cosx ^ cosx back

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