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keketsu Group Title

Would someone work through the derivative for this: (xy)^x = e Thanks.

  • 2 years ago
  • 2 years ago

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  1. TuringTest Group Title
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    logarithmic differentaition you want dy/dx I assume?

    • 2 years ago
  2. keketsu Group Title
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    yeah, please

    • 2 years ago
  3. TuringTest Group Title
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    \[(xy)^x=e\]\[x\ln(xy)=1\]\[x(\ln x+\ln y)=1\]taking the derivative gives\[\ln x+\ln y+x(\frac1x+\frac{y'}y)=0\]\[\ln(xy)+1+\frac xyy'=0\]\[y'=-\frac yx(\ln(xy)+1)\]now note that\[y=\frac{e^{1/x}}x\]so we get\[y'=-\frac{e^{1/x}}{x^2}(\ln(e^{1/x})+1)\]\[y'=-\frac{e^{1/x}}{x^2}(\frac1x+1)\]

    • 2 years ago
  4. keketsu Group Title
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    Wow, TuringTest, you're a wizard. Thanks. You've got two y derivatives then? Both would provide the same answer?

    • 2 years ago
  5. TuringTest Group Title
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    what do you mean two y derivatives?

    • 2 years ago
  6. keketsu Group Title
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    At the point you say "now note that," are you continuing to simplify, or approaching the problem in another way? I'm guessing by your response it's the first.

    • 2 years ago
  7. TuringTest Group Title
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    oh I just subbed in for y there... before that point the answer is still in terms of y. I assumed you wanted it all in terms of x, so I used the initial prompt to get y in terms of x\[(xy)^x=e\to xy=e^{1/x}\to y=\frac{e^{1/x}}x\]then I subbed that in

    • 2 years ago
  8. keketsu Group Title
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    Gotcha. I see it. And an even simpler question--how'd you take that first step? What identity did you use to get \[xln(xy)=1\]?

    • 2 years ago
  9. keketsu Group Title
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    I can see log x^c = c log x, but I'm confused about the e on the other side of the equation.

    • 2 years ago
  10. TuringTest Group Title
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    \[\ln e=1\]because that's the base of the logarithm

    • 2 years ago
  11. TuringTest Group Title
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    \[\log_aa=1\]in this case we have\[\ln e=\log_ee=1\]

    • 2 years ago
  12. TuringTest Group Title
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    if you don't remember that\[\large \log_aa=1\]you may want to review logarithms

    • 2 years ago
  13. keketsu Group Title
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    Haha, seriously. I'm attempting to take a comprehensive test over math I haven't studied in years. I have a reference manual with formulas and identities. When something isn't in there, I get a bit stuck. Not the optimal way to learn. But I appreciate you breaking this one down so thoroughly. Much appreciated.

    • 2 years ago
  14. keketsu Group Title
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    That is actually listed in my reference book. But I'm missing how you applied it. You didn't bring the e over ...

    • 2 years ago
  15. TuringTest Group Title
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    so you agree that \[\log_aa=1\]right?

    • 2 years ago
  16. keketsu Group Title
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    right

    • 2 years ago
  17. TuringTest Group Title
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    well the natural logarithm is just log base e\[\huge\ln e=\log_ee=1\]

    • 2 years ago
  18. TuringTest Group Title
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    (I just made that big to see the subscripts, sorry if it looks 'loud')

    • 2 years ago
  19. keketsu Group Title
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    No, I'm with you, I'm just missing how you apply that. If you take the left hand of the equation, using the following seems to work out the left side alone: \[\log x^c = c \log x\] How does the identity you're listing essentially turn e into 1?

    • 2 years ago
  20. TuringTest Group Title
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    I'm not sure I see the exact problem you're having do you now agree that\[\ln e=1\]or are you still doubtful?

    • 2 years ago
  21. keketsu Group Title
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    Ugh, I see it. You're applying ln to both sides. For some reason I was thinking the left-hand ln was coming from somewhere else. Got it! Thanks!

    • 2 years ago
  22. TuringTest Group Title
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    \[\large (xy)^x=e\]taking the natural log of both sides gives\[\ln(xy^x)=\ln e\]ah figured it out in the middle, good thanks for asking questions on what you didn't understand, that's a good thing you're welcome :D

    • 2 years ago
  23. keketsu Group Title
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    Ha, you bet. Many thanks for walking me through that.

    • 2 years ago
  24. keketsu Group Title
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    And go Chivas!

    • 2 years ago
  25. TuringTest Group Title
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    lol right on!

    • 2 years ago
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