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bettyboop8904

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

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  1. bettyboop8904
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    \[xe ^{y}+ye ^{x}=1\] at (0,1)

    • one year ago
  2. bettyboop8904
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    anybody??

    • one year ago
  3. bettyboop8904
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    @zepdrix

    • one year ago
  4. bettyboop8904
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    any idea how to do this one? = )

    • one year ago
  5. zepdrix
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    Bahhhhh gimme few minutes XD eating some food.

    • one year ago
  6. bettyboop8904
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    kk take your time = )

    • one year ago
  7. zepdrix
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    \[\huge xe^y+ye^x=1\]Were you able to take a derivative of this function? Or are we having some trouble on that part? :)

    • one year ago
  8. zepdrix
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    Or confused about the directions? heh

    • one year ago
  9. bettyboop8904
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    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: \[ye ^{x}=1-xe ^{y}\] \[y=\frac{ 1-xe ^{y} }{ e ^{x} }\] \[y'=\frac{ e ^{x}(y'xe ^{y})-(1-xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }\]

    • one year ago
  10. zepdrix
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    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:

    • one year ago
  11. bettyboop8904
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    hahaha ok so show me = )

    • one year ago
  12. zepdrix
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    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 :)

    • one year ago
  13. bettyboop8904
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    so is it everytime the derivative of y will be y*y' ?

    • one year ago
  14. zepdrix
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    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'.

    • one year ago
  15. zepdrix
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    \[\huge \color{salmon}{xe^y}+ye^x=1\]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,\[\large \color{salmon}{e^y+x(e^y)y'}\]

    • one year ago
  16. zepdrix
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    Understand how we did the first term? :)

    • one year ago
  17. bettyboop8904
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    ok that's right I remember now = )

    • one year ago
  18. bettyboop8904
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    that was for above hold on let me read what u posted lol

    • one year ago
  19. zepdrix
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    Hehe yah I understand :3

    • one year ago
  20. zepdrix
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    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.

    • one year ago
  21. bettyboop8904
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    ok so the next part would be (and altogether): \[(e ^{y}+xy'e ^{y})+(y'e ^{x}+ye ^{x})\]

    • one year ago
  22. bettyboop8904
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    =1 oops lol

    • one year ago
  23. zepdrix
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    Yes good c: `Except` we took the derivative of `both sides`, so what should the right side produce?

    • one year ago
  24. bettyboop8904
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    0

    • one year ago
  25. zepdrix
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    \[\huge e^y+xe ^y\color{orangered}{y'}+e^x\color{orangered}{y'}+ye^x=0\]Our goal from here is to solve for y'

    • one year ago
  26. zepdrix
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    Yes, good 0* blah that disappeared somehow lol

    • one year ago
  27. bettyboop8904
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    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

    • one year ago
  28. zepdrix
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    Understand how to do that? It's not too bad, it just requires a bit of nasty algebra :)

    • one year ago
  29. bettyboop8904
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    yeah give me a sec i love algebra = )

    • one year ago
  30. zepdrix
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    haha nice XD

    • one year ago
  31. zepdrix
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    Oh my goodness betty... -_- are you using the equation tool? zzzzzzz lol

    • one year ago
  32. bettyboop8904
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    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 \[xy'e ^{y}+y'e ^{x}=-e ^{y}-e ^{x}y\] \[y'(xe ^{y}+e ^{x})=-e ^{y}-e ^{x}y\] \[y'=\frac{ -e ^{y}-e ^{x}y }{ xe ^{y}+e ^{x} }\]

    • one year ago
  33. bettyboop8904
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    haha yes i was

    • one year ago
  34. zepdrix
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    looks good c:

    • one year ago
  35. bettyboop8904
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    ok so now I plug 0 in for x?

    • one year ago
  36. bettyboop8904
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    what is that lol jk

    • one year ago
  37. zepdrix
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    |dw:1359347099057:dw|

    • one year ago
  38. zepdrix
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    This isn't necessarily what the function looks like. It's just to give us an idea of what they're asking.

    • one year ago
  39. bettyboop8904
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    ok

    • one year ago
  40. zepdrix
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    They want us to form an equation for the `straight line` that is tangent to the function at x=0.

    • one year ago
  41. zepdrix
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    Remember the form of a straight line? :)

    • one year ago
  42. zepdrix
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    That's right! Your good ole buddy \(y=mx+b\). Remember him?

    • one year ago
  43. bettyboop8904
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    oh yeah ok i wasn't sure what u were asking lol

    • one year ago
  44. zepdrix
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    Our slope \(m\) is going to be equal to the slope of the tangent line at x=0. Meaning \(y'(0)=m\).

    • one year ago
  45. bettyboop8904
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    I was actually helping a person out today a lot with that form and the slope form with two points

    • one year ago
  46. zepdrix
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    Oh cool XD

    • one year ago
  47. zepdrix
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    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.

    • one year ago
  48. bettyboop8904
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    oh ok i was wondering what i would do with the y-values

    • one year ago
  49. bettyboop8904
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    i got \[\frac{ -e-1 }{ 1 }\]

    • one year ago
  50. zepdrix
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    Yep looks good.

    • one year ago
  51. zepdrix
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    \[\large y=mx+b \qquad \rightarrow \qquad y=-(e+1)x+b\] 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).

    • one year ago
  52. bettyboop8904
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    kk let me work it out = )

    • one year ago
  53. bettyboop8904
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    ok so this is what i got: \[y=-(e+1)x+1\] Now would you multiply it out like this: \[y= -ex-x+1\rightarrow -x(e+1)+1\]

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

    • one year ago
  55. bettyboop8904
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    and then thats it? = )

    • one year ago
  56. zepdrix
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    Yes, we found an equation for the line tangent to our curve at x=0, y=1.

    • one year ago
  57. bettyboop8904
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    awesome = ) ty

    • one year ago
  58. bettyboop8904
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    I got a couple others = ) stay posted lol

    • one year ago
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