anonymous
  • anonymous
????
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
So... derivative, ya? How we start? Chain rule?\[\Large\rm e^{xy}(xy)'=0\]Understand that step? :d what next?
anonymous
  • anonymous
So... derivative, ya? How we start? Chain rule?\[\Large\rm e^{xy}(xy)'=0\]Understand that step? :d what next?
anonymous
  • anonymous
Camila have we met before ?

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anonymous
  • anonymous
No I don't think so. I would first subtract 1 from the xy and put it in front of the e correct?
anonymous
  • anonymous
Where are you from?
zepdrix
  • zepdrix
Recall 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
  • zepdrix
So 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
  • anonymous
How would you derive the e^xy though? Because the product rule is f'(x)g(x)+f(x)g'(x)
zepdrix
  • zepdrix
So no, we're not apply power rule. We don't get an xy-1 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
  • anonymous
I think it would be 3e^x^2?
zepdrix
  • zepdrix
No, 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
  • zepdrix
\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(x^3)'\]\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(3x^2)\]
zepdrix
  • zepdrix
So 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
  • anonymous
Would you have to multiply it by its inverse?
zepdrix
  • zepdrix
You would multiply it by the `derivative of the inner function`. The inner function is the stuff up in the exponent. The xy.
zepdrix
  • zepdrix
\[\Large\rm \left(e^{xy}\right)'=e^{xy}\text{___}\]\[\Large\rm \left(e^{xy}\right)'=e^{xy}(xy)'\]So now we'll `setup` our product rule:\[\Large\rm \left(e^{xy}\right)'=e^{xy}(x'y+xy')\]
zepdrix
  • zepdrix
\[\large\rm x'=1\]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
  • zepdrix
What do you think lady Cam? :d too confusing?
anonymous
  • anonymous
I am very confused.
anonymous
  • anonymous
I know how to do the chain rule when it doesn't involve natural logs but now it got confusing.
zepdrix
  • zepdrix
We can actually use logs to simplify this problem BEFORE we take derivative. Maybe that would help a little bit.
zepdrix
  • zepdrix
So 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
  • zepdrix
But, if working with logs is really confusing, we can maybe avoid taking that route.
anonymous
  • anonymous
It 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
  • zepdrix
Woops, your derivative looks a little wonky.
zepdrix
  • zepdrix
First of all, let's deal with the right side. You forgot to take the derivative of the right side. (ln4)'=?
anonymous
  • anonymous
It equals 0
zepdrix
  • zepdrix
good good good, it's just a fancy looking constant. but it's still a constant. so derivative is 0 :)
zepdrix
  • zepdrix
\[\Large\rm xy=\ln4\]\[\Large\rm (xy)'=(\ln4)'\]\[\Large\rm (xy)'=0\]Now the left side... hmmm trying to figure out what you did there :d
zepdrix
  • zepdrix
I 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
  • zepdrix
x' 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
  • zepdrix
\[\Large\rm y+xy'=0\]So that finishes up the differentiation process at least, ya? :)
zepdrix
  • zepdrix
Now plug and chug
anonymous
  • anonymous
So its ln4+0=0 so it would just be ln4?
zepdrix
  • zepdrix
you're plugging in `1 for x`, and `ln4 for y`, i'm not sure where you're getting that 0 from :o
anonymous
  • anonymous
Because you said xy'
zepdrix
  • zepdrix
you're trying to solve for y', you don't plug anything into that.
zepdrix
  • zepdrix
\[\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0\]You're plugging your x and y values into these.
anonymous
  • anonymous
Ooooooo okay so it would be ln4+1=0 and then when you solve it would be -ln4+4?
zepdrix
  • zepdrix
\[\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0\]\[\Large\rm \color{orangered}{\ln4}+\color{orangered}{1}y'=0\]Your y' disappeared, I'm alil confused 0_o
anonymous
  • anonymous
Im 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
  • zepdrix
\[\Large\rm y'=-\ln4\]Yayyyy good job \c:/
zepdrix
  • zepdrix
\[\Large\rm \frac{dy}{dx}=-\ln 4\]
anonymous
  • anonymous
Thank you so much for helping me through this!!
zepdrix
  • zepdrix
But 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'
anonymous
  • anonymous
Oh okay

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