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

Thanks.

Question

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

turingtest · Answer

logarithmic differentaition

you want dy/dx I assume?

anonymous · Answer

yeah, please

turingtest · Answer

$$(xy)^x=e$$$$x\ln(xy)=1$$$$x(\ln x+\ln y)=1$$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$$y'=-\frac{e^{1/x}}{x^2}(\ln(e^{1/x})+1)$$$$y'=-\frac{e^{1/x}}{x^2}(\frac1x+1)$$

anonymous · Answer

Wow, TuringTest, you're a wizard.  Thanks.

You've got two y derivatives then?  Both would provide the same answer?

turingtest · Answer

what do you mean two y derivatives?

anonymous · Answer

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 · Answer

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 · Answer

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 · Answer

I can see log x^c = c log x, but I'm confused about the e on the other side of the equation.

turingtest · Answer

$$\ln e=1$$because that's the base of the logarithm

turingtest · Answer

$$\log_aa=1$$in this case we have$$\ln e=\log_ee=1$$

turingtest · Answer

if you don't remember that$$\large \log_aa=1$$you may want to review logarithms

anonymous · Answer

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 · Answer

That is actually listed in my reference book.  But I'm missing how you applied it.  You didn't bring the e over ...

turingtest · Answer

so you agree that $$\log_aa=1$$right?

anonymous · Answer

right

turingtest · Answer

well the natural logarithm is just log base e$$\huge\ln e=\log_ee=1$$

turingtest · Answer

(I just made that big to see the subscripts, sorry if it looks 'loud')

anonymous · Answer

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 · Answer

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 · Answer

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 · Answer

$$\large (xy)^x=e$$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 · Answer

Ha, you bet.  Many thanks for walking me through that.

anonymous · Answer

And go Chivas!

turingtest · Answer

lol right on!