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amorfide Group Title

Differentiate step by step I don't remember how to do it (4lnx-3)/(4lnx+3)

  • 2 years ago
  • 2 years ago

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  1. amorfide Group Title
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    would i do (vdu-udv)/v²

    • 2 years ago
  2. amorfide Group Title
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    @cwrw238

    • 2 years ago
  3. amorfide Group Title
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    @Hero can you help?

    • 2 years ago
  4. zepdrix Group Title
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    Are you allowed to simplify your answer before you differentiate? :) Because it's currently written as a pair of conjugates, if we multiply them together it'll simplify very nicely.

    • 2 years ago
  5. amorfide Group Title
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    no do it straight off use substitute v and u if needed

    • 2 years ago
  6. amorfide Group Title
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    actually i do not think method matters xD

    • 2 years ago
  7. zepdrix Group Title
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    Ok :) In that case, it looks like we have the product of two functions involving x. So we'll need to apply the product rule. The one that you posted above is the quotient rule :O we don't want to you that formula. \[\huge (uv)'=u'v+uv'\] This is the one we want :)

    • 2 years ago
  8. amorfide Group Title
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    i am sure it is quotient,...

    • 2 years ago
  9. zepdrix Group Title
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    Oh i didn't see the division symbol in there sorry, it was hidden for some reason lolol

    • 2 years ago
  10. amorfide Group Title
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    i have the answer no method answer is 24/x(4lnx+3)²

    • 2 years ago
  11. zepdrix Group Title
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    I hate that tiny text :)

    • 2 years ago
  12. zepdrix Group Title
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    k you have the right formula then, just need help setting it up? :o

    • 2 years ago
  13. amorfide Group Title
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    well i know which is v and u i dont know the differential of 4lnx -3

    • 2 years ago
  14. zepdrix Group Title
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    Recall this:\[(\ln x)'=\frac{ 1 }{ x }\] So let's apply that to our u. \[\large u=4\ln x - 3\]\[\large u'=4\frac{ 1 }{ x } - 0\] The 4 is just a constant, so we can ignore that while we differentiate.

    • 2 years ago
  15. amorfide Group Title
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    I feel dumb, thank you <3

    • 2 years ago
  16. zepdrix Group Title
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    heh :3

    • 2 years ago
  17. amorfide Group Title
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    well.. i get |dw:1352060900067:dw|

    • 2 years ago
  18. amorfide Group Title
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    if i multiply by x/x will that give me the right answer?

    • 2 years ago
  19. amorfide Group Title
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    @zepdrix

    • 2 years ago
  20. amorfide Group Title
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    so how would i find the gradient of (4lnx-3)/(4lnx+3)

    • 2 years ago
  21. zepdrix Group Title
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    |dw:1352061081956:dw| Hmmm I'm confused how you get 24/x on the top :O

    • 2 years ago
  22. zepdrix Group Title
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    Nevermind I see it now XD

    • 2 years ago
  23. zepdrix Group Title
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    Me so slow today :3

    • 2 years ago
  24. amorfide Group Title
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    i moved on, i need the gradient lol... how?

    • 2 years ago
  25. zepdrix Group Title
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    the gradient? :( mmm I'm in the US, is that what we call the slope i guess? :O

    • 2 years ago
  26. amorfide Group Title
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    yeah slope

    • 2 years ago
  27. zepdrix Group Title
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    Are you trying to form an equation for a particular line tangent to the function? :O We need a specific point along the function to start with if that's the case.

    • 2 years ago
  28. amorfide Group Title
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    crosses the x axis y=(4lnx-3)/(4lnx+3) get the exact value of the gradient

    • 2 years ago
  29. zepdrix Group Title
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    |dw:1352061602252:dw| Understand how we got that particular x value? It's a little bit tricky near the end there.

    • 2 years ago
  30. amorfide Group Title
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    ahh thank you! forget to make it equal to zero my bad xD

    • 2 years ago
  31. zepdrix Group Title
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    Understand how to find the value of the gradient from there? :) Plugging that point into the derivative function and such? :D

    • 2 years ago
  32. amorfide Group Title
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    36e^3/4 ?

    • 2 years ago
  33. amorfide Group Title
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    ahh that is just the denominator my bad (2/3)e^(3/4)

    • 2 years ago
  34. amorfide Group Title
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    that should be -3/4

    • 2 years ago
  35. zepdrix Group Title
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    |dw:1352062210018:dw| Hmm yah that looks right, good job! :)

    • 2 years ago
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