## bettyboop8904 2 years ago Can anyone help me figure this problem out pretty please? Find an equation of the tangent line to the curve xe^(y)+ye^(x)=1 at the point (0,1) The equation is rewritten in the post with the equation converter helper = )

1. bettyboop8904

$xe ^{y}+ye ^{x}=1$ at (0,1)

2. bettyboop8904

anybody??

3. bettyboop8904

@zepdrix

4. bettyboop8904

any idea how to do this one? = )

5. zepdrix

Bahhhhh gimme few minutes XD eating some food.

6. bettyboop8904

kk take your time = )

7. zepdrix

$\huge xe^y+ye^x=1$Were you able to take a derivative of this function? Or are we having some trouble on that part? :)

8. zepdrix

Or confused about the directions? heh

9. bettyboop8904

I was a little confused, since my teacher added implicit differentiation to our course last semester I always get confused. This is what I got when I took the derivative but I'm not sure if its correct: $ye ^{x}=1-xe ^{y}$ $y=\frac{ 1-xe ^{y} }{ e ^{x} }$ $y'=\frac{ e ^{x}(y'xe ^{y})-(1-xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }$

10. zepdrix

ew ew ew don't try to turn it into an explicit function in $$x$$. It's actually impossible in this case. So just leave it alone, and take the derivative from where it is at the start D:

11. bettyboop8904

hahaha ok so show me = )

12. zepdrix

Implicit isn't too bad, just remember that whenever you take the derivative of a y term, a y' will pop out. Ok ok ok :)

13. bettyboop8904

so is it everytime the derivative of y will be y*y' ?

14. zepdrix

Everything you take the derivative of y, you'll apply the power rule just like you would for x. Giving you 1 (not y), and then you attach a y' to it. So the derivative of y will be 1*y'.

15. zepdrix

$\huge \color{salmon}{xe^y}+ye^x=1$We'll take the derivative of the pink term first. Applying the Product Rule will give us,$\large \color{salmon}{(x)'e^y+x(e^y)'}$Which simplifies to,$\large \color{salmon}{e^y+x(e^y)y'}$

16. zepdrix

Understand how we did the first term? :)

17. bettyboop8904

ok that's right I remember now = )

18. bettyboop8904

that was for above hold on let me read what u posted lol

19. zepdrix

Hehe yah I understand :3

20. zepdrix

On the second part, we had to apply the chain rule. e^y gave us e^y. But then we also have to multiply by the derivative of the exponent.

21. bettyboop8904

ok so the next part would be (and altogether): $(e ^{y}+xy'e ^{y})+(y'e ^{x}+ye ^{x})$

22. bettyboop8904

=1 oops lol

23. zepdrix

Yes good c: Except we took the derivative of both sides, so what should the right side produce?

24. bettyboop8904

0

25. zepdrix

$\huge e^y+xe ^y\color{orangered}{y'}+e^x\color{orangered}{y'}+ye^x=0$Our goal from here is to solve for y'

26. zepdrix

Yes, good 0* blah that disappeared somehow lol

27. bettyboop8904

haha its ok i was trying to figure out what to do next lol i was going to say plug 0 in for x but solve for y' that makes sense lol

28. zepdrix

Understand how to do that? It's not too bad, it just requires a bit of nasty algebra :)

29. bettyboop8904

yeah give me a sec i love algebra = )

30. zepdrix

haha nice XD

31. zepdrix

Oh my goodness betty... -_- are you using the equation tool? zzzzzzz lol

32. bettyboop8904

Ok I do love algebra but I could be wrong with this one but I think I might be right at the same time lol $xy'e ^{y}+y'e ^{x}=-e ^{y}-e ^{x}y$ $y'(xe ^{y}+e ^{x})=-e ^{y}-e ^{x}y$ $y'=\frac{ -e ^{y}-e ^{x}y }{ xe ^{y}+e ^{x} }$

33. bettyboop8904

haha yes i was

34. zepdrix

looks good c:

35. bettyboop8904

ok so now I plug 0 in for x?

36. bettyboop8904

what is that lol jk

37. zepdrix

|dw:1359347099057:dw|

38. zepdrix

This isn't necessarily what the function looks like. It's just to give us an idea of what they're asking.

39. bettyboop8904

ok

40. zepdrix

They want us to form an equation for the straight line that is tangent to the function at x=0.

41. zepdrix

Remember the form of a straight line? :)

42. zepdrix

That's right! Your good ole buddy $$y=mx+b$$. Remember him?

43. bettyboop8904

oh yeah ok i wasn't sure what u were asking lol

44. zepdrix

Our slope $$m$$ is going to be equal to the slope of the tangent line at x=0. Meaning $$y'(0)=m$$.

45. bettyboop8904

I was actually helping a person out today a lot with that form and the slope form with two points

46. zepdrix

Oh cool XD

47. zepdrix

I guess I should have written it like this,$y'(0,1)=m$We're going to plug in the given x and y value to get our m.

48. bettyboop8904

oh ok i was wondering what i would do with the y-values

49. bettyboop8904

i got $\frac{ -e-1 }{ 1 }$

50. zepdrix

Yep looks good.

51. zepdrix

$\large y=mx+b \qquad \rightarrow \qquad y=-(e+1)x+b$ And now we only need to find the y-intercept b. To do so, we'll again use the point that ws given (0,1).

52. bettyboop8904

kk let me work it out = )

53. bettyboop8904

ok so this is what i got: $y=-(e+1)x+1$ Now would you multiply it out like this: $y= -ex-x+1\rightarrow -x(e+1)+1$

54. zepdrix

No don't multiply it out c: looks good! Hopefully we didn't make any silly mistakes in there. Looks correct though :D

55. bettyboop8904

and then thats it? = )

56. zepdrix

Yes, we found an equation for the line tangent to our curve at x=0, y=1.

57. bettyboop8904

awesome = ) ty

58. bettyboop8904

I got a couple others = ) stay posted lol