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Dodo1

  • 2 years ago

f(x)=[Inx]^4

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  1. geerky42
    • 2 years ago
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    Do you know chain rule? @Dodo1

  2. Dodo1
    • 2 years ago
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    Logarithmic Derivative rules that i have to use

  3. Dodo1
    • 2 years ago
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    Yes I do, f'(g(x).g(x)'

  4. geerky42
    • 2 years ago
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    Yeah. Use it. You should get 4(lnx)³ · (1/x)

  5. Dodo1
    • 2 years ago
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    is that answer?

  6. Dodo1
    • 2 years ago
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    its not Logarithmic Derivative is it?

  7. Dodo1
    • 2 years ago
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    @zepdrix any ideas?

  8. zepdrix
    • 2 years ago
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    Why would they want you to apply Logarithmic Differentiation to this problem? Very strange... Ok I guess we can do it though :\ Rewrite \(\large f(x)\) as \(\large y\). Then take the log (base e) of both sides. \[\large \ln y=\ln\left[(\ln x)^4\right]\]

  9. Dodo1
    • 2 years ago
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    put In to both side, ok

  10. zepdrix
    • 2 years ago
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    Using a rule of logarithms,\[\large \color{royalblue}{\log(b^\color{orangered}{a})=\color{orangered}{a}\log(b)}\] We can bring that 4 out front.

  11. zepdrix
    • 2 years ago
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    This is going to be a much more difficult problem using Logarithmic Differentiation. I'm not sure why the directions would suggest you do this :) lol

  12. zepdrix
    • 2 years ago
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    \[\large \ln y=4 \ln\left[\ln x\right]\]

  13. zepdrix
    • 2 years ago
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    Take the derivative of both sides with respect to x. Wanna take a shot at that part? :) What do you get on the left side of the equation?

  14. Dodo1
    • 2 years ago
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    4log(x)?

  15. zepdrix
    • 2 years ago
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    whut? D:

  16. geerky42
    • 2 years ago
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    What do you get on the left side of the equation? LEFT side

  17. Dodo1
    • 2 years ago
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    oh left x/1?

  18. Dodo1
    • 2 years ago
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    No its just Y?

  19. Dodo1
    • 2 years ago
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    Im really bad at log stff!

  20. Dodo1
    • 2 years ago
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    @geerky42 ?

  21. Dodo1
    • 2 years ago
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    Oh! I see ops

  22. zepdrix
    • 2 years ago
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    y'/y

  23. Dodo1
    • 2 years ago
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    1/y' is y'/y?

  24. zepdrix
    • 2 years ago
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    \[\large \frac{d}{dx}\ln y \qquad = \qquad \frac{1}{y}\cdot\frac{dy}{dx}\]

  25. Dodo1
    • 2 years ago
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    thats cool!

  26. Dodo1
    • 2 years ago
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    so 1/y*dy/x= In[Inx]^4. do i use chain rule for the next step?

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