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bettyboop8904

  • one year 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 = )

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

  2. bettyboop8904
    • one year ago
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    anybody??

  3. bettyboop8904
    • one year ago
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    @zepdrix

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

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

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

  7. zepdrix
    • one year ago
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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? :)

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

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

  10. zepdrix
    • one year ago
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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:

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

  12. zepdrix
    • one year ago
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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 :)

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

  14. zepdrix
    • one year ago
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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'.

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

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

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

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

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

  20. zepdrix
    • one year ago
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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.

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

  22. bettyboop8904
    • one year ago
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    =1 oops lol

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

  24. bettyboop8904
    • one year ago
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    0

  25. zepdrix
    • one year ago
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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'

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

  27. bettyboop8904
    • one year ago
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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

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

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

  30. zepdrix
    • one year ago
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    haha nice XD

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

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

  33. bettyboop8904
    • one year ago
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    haha yes i was

  34. zepdrix
    • one year ago
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    looks good c:

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

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

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

  38. zepdrix
    • one year ago
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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.

  39. bettyboop8904
    • one year ago
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    ok

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

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

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

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

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

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

  46. zepdrix
    • one year ago
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    Oh cool XD

  47. zepdrix
    • one year ago
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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.

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

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

  50. zepdrix
    • one year ago
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    Yep looks good.

  51. zepdrix
    • one year ago
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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).

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

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

  54. zepdrix
    • one year ago
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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

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

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

  57. bettyboop8904
    • one year ago
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    awesome = ) ty

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

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