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lgbasallote

LGBADERIVATIVE: \[\huge y = e^{k\tan \sqrt x}\]

  • one year ago
  • one year ago

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  1. lgbasallote
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    im guessing ln again

    • one year ago
  2. freckles
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    Well you can use log diff like I mention before Or you can use straight chain man

    • one year ago
  3. lgbasallote
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    \[\ln y = k\tan \sqrt x\] now what...

    • one year ago
  4. freckles
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    \[y=e^{f(x)} => y'=f'(x)e^{f(x)}\]

    • one year ago
  5. lgbasallote
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    i uess i make it \[\ln y = k\frac{\sin \sqrt x}{\cos \sqrt x}\]

    • one year ago
  6. lgbasallote
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    this seems complicated :|

    • one year ago
  7. freckles
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    \[y=e^{f(x)}\] \[\ln(y)=\ln(e^{f(x)})\] \[\ln(y)=f(x) \cdot \ln(e)\] \[\ln(y)=f(x) (1)\] \[\ln(y)=f(x)\] \[\frac{y'}{y}=f'(x)\] \[y'=y f'(x)\] \[y'=e^{f(x)} f'(x)\]

    • one year ago
  8. lgbasallote
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    so my problem is that f'(x) now..

    • one year ago
  9. Zarkon
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    \[\frac{d}{dx}\tan(u)=\sec^2(u)\frac{du}{dx}\]

    • one year ago
  10. freckles
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    ^ what zarkon said :)

    • one year ago
  11. lgbasallote
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    uhmm oh yeah...how could i forget *facepalm*

    • one year ago
  12. lgbasallote
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    i need to take a rest hahahaha so it will be \[y' = \frac{e^{k\tan \sqrt x} k\sec^2 \sqrt x}{2\sqrt x}\] right?

    • one year ago
  13. freckles
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    Yes that looks good to me I see no error

    • one year ago
  14. lgbasallote
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    thanks :DDD

    • one year ago
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