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anonymous
 one year ago
????
anonymous
 one year ago
????

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So... derivative, ya? How we start? Chain rule?\[\Large\rm e^{xy}(xy)'=0\]Understand that step? :d what next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So... derivative, ya? How we start? Chain rule?\[\Large\rm e^{xy}(xy)'=0\]Understand that step? :d what next?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Camila have we met before ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No I don't think so. I would first subtract 1 from the xy and put it in front of the e correct?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Recall your derivative for an exponential function,\[\Large\rm y=e^{u}\]\[\Large\rm y'=e^{u}u'\]Example:\[\Large\rm y=e^{2x}\qquad\to\qquad y'=e^{2x}(2x)'\]\[\Large\rm y'=e^{2x}(2)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So for our problem, I started out by taking the derivative of both sides,\[\Large\rm (e^{xy})'=(4)'\]\[\Large\rm e^{xy}(xy)'=0\]And from here, to finish up the differentiation procress, we need to apply our product rule. See how we have x and y multiplying? Ya product rule :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would you derive the e^xy though? Because the product rule is f'(x)g(x)+f(x)g'(x)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So no, we're not apply power rule. We don't get an xy1 or anything like that :) We're applying exponential rule. Exponential gives us the `same thing back` when we differentiate, just gotta remember to chain rule. Like do you understand how to take this derivative? :o\[\Large\rm \left(e^{x^3}\right)'\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it would be 3e^x^2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4No, we should get the `same thing back` when we differentiate the exponential.\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}\text{__}\]See? Same thing as a result. But we need to also chain rule: multiply by the derivative of the inner function. The inner function being x^3 in this case.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(x^3)'\]\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(3x^2)\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So back to our problem:\[\Large\rm \left(e^{xy}\right)'=e^{xy}\text{___}\]We get the same thing back... but we're missing something, have to apply our chain rule.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would you have to multiply it by its inverse?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4You would multiply it by the `derivative of the inner function`. The inner function is the stuff up in the exponent. The xy.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm \left(e^{xy}\right)'=e^{xy}\text{___}\]\[\Large\rm \left(e^{xy}\right)'=e^{xy}(xy)'\]So now we'll `setup` our product rule:\[\Large\rm \left(e^{xy}\right)'=e^{xy}(x'y+xy')\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\large\rm x'=1\]Derivative of x, with respect to x, is simply 1, ya?\[\Large\rm \left(e^{xy}\right)'=e^{xy}(x'y+xy')\]\[\Large\rm \left(e^{xy}\right)'=e^{xy}(1\cdot y+xy')\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4What do you think lady Cam? :d too confusing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know how to do the chain rule when it doesn't involve natural logs but now it got confusing.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4We can actually use logs to simplify this problem BEFORE we take derivative. Maybe that would help a little bit.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4So like... if you take the natural log of each side at the start,\[\Large\rm \ln\left[e^{xy}\right]=\ln(4)\]Recall your log property that looks something like this,\[\rm \log(a^b)=b \log(a)\]The exponent can come out in front as a multiplier. So for our problem, we apply that rule and get,\[\Large\rm (xy) \ln[e]=\ln(4)\]ln(e) will simplify to 1, (check that with your calculator if you need). So we have:\[\Large\rm xy=\ln4\]This is the SAME PROBLEM. It just might be easier to work with from here.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4But, if working with logs is really confusing, we can maybe avoid taking that route.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It might be easier. So from xy=ln4 could I use product rule to get x+y=ln4 and then plug in my given values?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4Woops, your derivative looks a little wonky.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4First of all, let's deal with the right side. You forgot to take the derivative of the right side. (ln4)'=?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4good good good, it's just a fancy looking constant. but it's still a constant. so derivative is 0 :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm xy=\ln4\]\[\Large\rm (xy)'=(\ln4)'\]\[\Large\rm (xy)'=0\]Now the left side... hmmm trying to figure out what you did there :d

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4I would recommend that you take the steps to first SETUP your product rule. Don't just plug the stuff in until you're comfortable with it.\[\Large\rm (fg)'=f'g+fg'\]Yes? So for our problem:\[\Large\rm (xy)'=x'y+xy'\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4x' is just the derivative of x, which is simply 1.\[\Large\rm (xy)'=y+xy'\]But we cannot treat y' in the same way. This is our derivative term, dy/dx.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm y+xy'=0\]So that finishes up the differentiation process at least, ya? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So its ln4+0=0 so it would just be ln4?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4you're plugging in `1 for x`, and `ln4 for y`, i'm not sure where you're getting that 0 from :o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because you said xy'

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4you're trying to solve for y', you don't plug anything into that.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0\]You're plugging your x and y values into these.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ooooooo okay so it would be ln4+1=0 and then when you solve it would be ln4+4?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0\]\[\Large\rm \color{orangered}{\ln4}+\color{orangered}{1}y'=0\]Your y' disappeared, I'm alil confused 0_o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im so sorry I get really confused with math. Okay so you divided 0 by 1 and get 0 and subtract ln4 to get ln4?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm y'=\ln4\]Yayyyy good job \c:/

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large\rm \frac{dy}{dx}=\ln 4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for helping me through this!!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.4But in your final steps there, don't be silly. No reason to divide by 1 :) Just ignore the 1. 1y' is the same as y'
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