## anonymous one year ago ????

1. zepdrix

So... derivative, ya? How we start? Chain rule?$\Large\rm e^{xy}(xy)'=0$Understand that step? :d what next?

2. anonymous

So... derivative, ya? How we start? Chain rule?$\Large\rm e^{xy}(xy)'=0$Understand that step? :d what next?

3. anonymous

Camila have we met before ?

4. anonymous

No I don't think so. I would first subtract 1 from the xy and put it in front of the e correct?

5. anonymous

Where are you from?

6. 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)$

7. 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

8. anonymous

How would you derive the e^xy though? Because the product rule is f'(x)g(x)+f(x)g'(x)

9. 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$\Large\rm \left(e^{x^3}\right)'$

10. anonymous

I think it would be 3e^x^2?

11. 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.

12. 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)$

13. 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.

14. anonymous

Would you have to multiply it by its inverse?

15. zepdrix

You would multiply it by the derivative of the inner function. The inner function is the stuff up in the exponent. The xy.

16. 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')$

17. 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')$

18. zepdrix

What do you think lady Cam? :d too confusing?

19. anonymous

I am very confused.

20. anonymous

I know how to do the chain rule when it doesn't involve natural logs but now it got confusing.

21. zepdrix

We can actually use logs to simplify this problem BEFORE we take derivative. Maybe that would help a little bit.

22. 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.

23. zepdrix

But, if working with logs is really confusing, we can maybe avoid taking that route.

24. 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?

25. zepdrix

Woops, your derivative looks a little wonky.

26. zepdrix

First of all, let's deal with the right side. You forgot to take the derivative of the right side. (ln4)'=?

27. anonymous

It equals 0

28. zepdrix

good good good, it's just a fancy looking constant. but it's still a constant. so derivative is 0 :)

29. 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

30. 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:$\Large\rm (xy)'=x'y+xy'$

31. 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.

32. zepdrix

$\Large\rm y+xy'=0$So that finishes up the differentiation process at least, ya? :)

33. zepdrix

Now plug and chug

34. anonymous

So its ln4+0=0 so it would just be ln4?

35. zepdrix

you're plugging in 1 for x, and ln4 for y, i'm not sure where you're getting that 0 from :o

36. anonymous

Because you said xy'

37. zepdrix

you're trying to solve for y', you don't plug anything into that.

38. zepdrix

$\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0$You're plugging your x and y values into these.

39. anonymous

Ooooooo okay so it would be ln4+1=0 and then when you solve it would be -ln4+4?

40. 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

41. 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?

42. zepdrix

$\Large\rm y'=-\ln4$Yayyyy good job \c:/

43. zepdrix

$\Large\rm \frac{dy}{dx}=-\ln 4$

44. anonymous

Thank you so much for helping me through this!!

45. 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'

46. anonymous

Oh okay