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cunninnc Group Title

find the derivative of y=log10(5x^4-2/x^3)

  • 2 years ago
  • 2 years ago

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  1. MrMoose Group Title
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    this involves both the chain rule and the quotient rule

    • 2 years ago
  2. Mimi_x3 Group Title
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    or you can use log laws to change it first; then diferentiate it, would be easier.

    • 2 years ago
  3. cunninnc Group Title
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    @failmathmajor only the 2/x^3 is a fraction

    • 2 years ago
  4. Mimi_x3 Group Title
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    is it \[\log\left(5x^{4}-\frac{2}{x^{3}}\right) ?\]

    • 2 years ago
  5. MrMoose Group Title
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    that would be much easier

    • 2 years ago
  6. MrMoose Group Title
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    so using log properties you can say that \[\log_{10} {x} = \ln x / \ln 10 \]

    • 2 years ago
  7. MrMoose Group Title
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    ln10 is a constant, you can factor that out

    • 2 years ago
  8. MrMoose Group Title
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    can you use the chain rule fluently?

    • 2 years ago
  9. cunninnc Group Title
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    y'=pu(x) * u'(x) right?

    • 2 years ago
  10. MrMoose Group Title
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    d(f(g(x))/dx = (dg(x)/dx)*(df(g(x))/dx) is the way I know it

    • 2 years ago
  11. Mimi_x3 Group Title
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    Or..\[y= \log\left(5x^{4}-\frac{2}{x^{3}}\right) y = \log\left(\frac{5x^{7}-2}{x^{3}}\right) => y = \log\left(5x^{7}-2\right) - \log(x^3)\] Might be easier; than the quotient role and chain rule..

    • 2 years ago
  12. MrMoose Group Title
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    that may look confusing now that I see it

    • 2 years ago
  13. MrMoose Group Title
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    don't need the quotient in this case: can treat /x^3 as a x^-3

    • 2 years ago
  14. Mimi_x3 Group Title
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    @cunninnc: are you able to do it now?

    • 2 years ago
  15. MrMoose Group Title
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    and you would still need the chain rule for that anyway

    • 2 years ago
  16. MrMoose Group Title
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    but enough arguing

    • 2 years ago
  17. Mimi_x3 Group Title
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    Did an argument started? lol

    • 2 years ago
  18. MrMoose Group Title
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    \[(d(\ln (5x^4 - 2x^{-3}))/dx )/ \ln10\] is what it simplifies down to, to be concise

    • 2 years ago
  19. cunninnc Group Title
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    @Mimi_x3 kinda .... i see mr. moose got 2x^-3 where does ^-3 come froms

    • 2 years ago
  20. MrMoose Group Title
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    \[2/x^3= 2 * x^{-3}\]

    • 2 years ago
  21. Mimi_x3 Group Title
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    Sorry i don't know what MrMoose is doing. @MrMoose: Use \frac{x}{y} for fractions :)

    • 2 years ago
  22. Mimi_x3 Group Title
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    Or why not try the method that i used :) \[ \frac{d}{dx} \log\left(5x^{7}-2\right) -\frac{d}{dx} \log(x^3)\]

    • 2 years ago
  23. MrMoose Group Title
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    when you divide you subtract exponents, so that is equivalent to saying \[2 * \frac{x^0}{x^3}\] then subtract exponents in division

    • 2 years ago
  24. MrMoose Group Title
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    \[\frac{\frac{d(\ln(5x^4−2x^{−3}))}{dx}}{\ln10}\]

    • 2 years ago
  25. MrMoose Group Title
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    @catamountz15 what?

    • 2 years ago
  26. MrMoose Group Title
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    I am almost entirely sure that that isn't a form of the answer.

    • 2 years ago
  27. catamountz15 Group Title
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    Here are the steps in to solving this problem.

    • 2 years ago
  28. MrMoose Group Title
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    that isn't what you wrote though

    • 2 years ago
  29. catamountz15 Group Title
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    My apologies that was an answer to a different problem.

    • 2 years ago
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