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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2logarithmic differentaition you want dy/dx I assume?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2\[(xy)^x=e\]\[x\ln(xy)=1\]\[x(\ln x+\ln y)=1\]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\[y'=\frac{e^{1/x}}{x^2}(\ln(e^{1/x})+1)\]\[y'=\frac{e^{1/x}}{x^2}(\frac1x+1)\]

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2Wow, TuringTest, you're a wizard. Thanks. You've got two y derivatives then? Both would provide the same answer?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2what do you mean two y derivatives?

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2At 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
 3 years ago
Best ResponseYou've already chosen the best response.2oh 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

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2Gotcha. I see it. And an even simpler questionhow'd you take that first step? What identity did you use to get \[xln(xy)=1\]?

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2I can see log x^c = c log x, but I'm confused about the e on the other side of the equation.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2\[\ln e=1\]because that's the base of the logarithm

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2\[\log_aa=1\]in this case we have\[\ln e=\log_ee=1\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2if you don't remember that\[\large \log_aa=1\]you may want to review logarithms

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2Haha, 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.

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2That is actually listed in my reference book. But I'm missing how you applied it. You didn't bring the e over ...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2so you agree that \[\log_aa=1\]right?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2well the natural logarithm is just log base e\[\huge\ln e=\log_ee=1\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2(I just made that big to see the subscripts, sorry if it looks 'loud')

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2No, 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
 3 years ago
Best ResponseYou've already chosen the best response.2I'm not sure I see the exact problem you're having do you now agree that\[\ln e=1\]or are you still doubtful?

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2Ugh, I see it. You're applying ln to both sides. For some reason I was thinking the lefthand ln was coming from somewhere else. Got it! Thanks!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large (xy)^x=e\]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

keketsu
 3 years ago
Best ResponseYou've already chosen the best response.2Ha, you bet. Many thanks for walking me through that.
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