anonymous
  • anonymous
Would someone work through the derivative for this: (xy)^x = e Thanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
logarithmic differentaition you want dy/dx I assume?
anonymous
  • anonymous
yeah, please
TuringTest
  • TuringTest
\[(xy)^x=e\]\[x\ln(xy)=1\]\[x(\ln x+\ln y)=1\]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\[y'=-\frac{e^{1/x}}{x^2}(\ln(e^{1/x})+1)\]\[y'=-\frac{e^{1/x}}{x^2}(\frac1x+1)\]

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anonymous
  • anonymous
Wow, TuringTest, you're a wizard. Thanks. You've got two y derivatives then? Both would provide the same answer?
TuringTest
  • TuringTest
what do you mean two y derivatives?
anonymous
  • anonymous
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.
TuringTest
  • TuringTest
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
anonymous
  • anonymous
Gotcha. I see it. And an even simpler question--how'd you take that first step? What identity did you use to get \[xln(xy)=1\]?
anonymous
  • anonymous
I can see log x^c = c log x, but I'm confused about the e on the other side of the equation.
TuringTest
  • TuringTest
\[\ln e=1\]because that's the base of the logarithm
TuringTest
  • TuringTest
\[\log_aa=1\]in this case we have\[\ln e=\log_ee=1\]
TuringTest
  • TuringTest
if you don't remember that\[\large \log_aa=1\]you may want to review logarithms
anonymous
  • anonymous
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.
anonymous
  • anonymous
That is actually listed in my reference book. But I'm missing how you applied it. You didn't bring the e over ...
TuringTest
  • TuringTest
so you agree that \[\log_aa=1\]right?
anonymous
  • anonymous
right
TuringTest
  • TuringTest
well the natural logarithm is just log base e\[\huge\ln e=\log_ee=1\]
TuringTest
  • TuringTest
(I just made that big to see the subscripts, sorry if it looks 'loud')
anonymous
  • anonymous
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?
TuringTest
  • TuringTest
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?
anonymous
  • anonymous
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!
TuringTest
  • TuringTest
\[\large (xy)^x=e\]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
anonymous
  • anonymous
Ha, you bet. Many thanks for walking me through that.
anonymous
  • anonymous
And go Chivas!
TuringTest
  • TuringTest
lol right on!

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