## bettyboop8904 Group Title 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 = ) one year ago one year ago

1. bettyboop8904 Group Title

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

2. bettyboop8904 Group Title

anybody??

3. bettyboop8904 Group Title

@zepdrix

4. bettyboop8904 Group Title

any idea how to do this one? = )

5. zepdrix Group Title

Bahhhhh gimme few minutes XD eating some food.

6. bettyboop8904 Group Title

kk take your time = )

7. zepdrix Group Title

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

8. zepdrix Group Title

Or confused about the directions? heh

9. bettyboop8904 Group Title

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 Group Title

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 Group Title

hahaha ok so show me = )

12. zepdrix Group Title

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 Group Title

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

14. zepdrix Group Title

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 Group Title

$\huge \color{salmon}{xe^y}+ye^x=1$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 Group Title

Understand how we did the first term? :)

17. bettyboop8904 Group Title

ok that's right I remember now = )

18. bettyboop8904 Group Title

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

19. zepdrix Group Title

Hehe yah I understand :3

20. zepdrix Group Title

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 Group Title

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

22. bettyboop8904 Group Title

=1 oops lol

23. zepdrix Group Title

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

24. bettyboop8904 Group Title

0

25. zepdrix Group Title

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

26. zepdrix Group Title

Yes, good 0* blah that disappeared somehow lol

27. bettyboop8904 Group Title

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 Group Title

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

29. bettyboop8904 Group Title

yeah give me a sec i love algebra = )

30. zepdrix Group Title

haha nice XD

31. zepdrix Group Title

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

32. bettyboop8904 Group Title

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 Group Title

haha yes i was

34. zepdrix Group Title

looks good c:

35. bettyboop8904 Group Title

ok so now I plug 0 in for x?

36. bettyboop8904 Group Title

what is that lol jk

37. zepdrix Group Title

|dw:1359347099057:dw|

38. zepdrix Group Title

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

39. bettyboop8904 Group Title

ok

40. zepdrix Group Title

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

41. zepdrix Group Title

Remember the form of a straight line? :)

42. zepdrix Group Title

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

43. bettyboop8904 Group Title

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

44. zepdrix Group Title

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

45. bettyboop8904 Group Title

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

46. zepdrix Group Title

Oh cool XD

47. zepdrix Group Title

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 Group Title

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

49. bettyboop8904 Group Title

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

50. zepdrix Group Title

Yep looks good.

51. zepdrix Group Title

$\large y=mx+b \qquad \rightarrow \qquad y=-(e+1)x+b$ 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 Group Title

kk let me work it out = )

53. bettyboop8904 Group Title

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 Group Title

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 Group Title

and then thats it? = )

56. zepdrix Group Title

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

57. bettyboop8904 Group Title

awesome = ) ty

58. bettyboop8904 Group Title

I got a couple others = ) stay posted lol