anonymous
  • anonymous
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 = )
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[xe ^{y}+ye ^{x}=1\] at (0,1)
anonymous
  • anonymous
anybody??
anonymous
  • anonymous
@zepdrix

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anonymous
  • anonymous
any idea how to do this one? = )
zepdrix
  • zepdrix
Bahhhhh gimme few minutes XD eating some food.
anonymous
  • anonymous
kk take your time = )
zepdrix
  • 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? :)
zepdrix
  • zepdrix
Or confused about the directions? heh
anonymous
  • anonymous
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: \[ye ^{x}=1-xe ^{y}\] \[y=\frac{ 1-xe ^{y} }{ e ^{x} }\] \[y'=\frac{ e ^{x}(y'xe ^{y})-(1-xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }\]
zepdrix
  • 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:
anonymous
  • anonymous
hahaha ok so show me = )
zepdrix
  • 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 :)
anonymous
  • anonymous
so is it everytime the derivative of y will be y*y' ?
zepdrix
  • 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'.
zepdrix
  • 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,\[\large \color{salmon}{e^y+x(e^y)y'}\]
zepdrix
  • zepdrix
Understand how we did the first term? :)
anonymous
  • anonymous
ok that's right I remember now = )
anonymous
  • anonymous
that was for above hold on let me read what u posted lol
zepdrix
  • zepdrix
Hehe yah I understand :3
zepdrix
  • 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.
anonymous
  • anonymous
ok so the next part would be (and altogether): \[(e ^{y}+xy'e ^{y})+(y'e ^{x}+ye ^{x})\]
anonymous
  • anonymous
=1 oops lol
zepdrix
  • zepdrix
Yes good c: `Except` we took the derivative of `both sides`, so what should the right side produce?
anonymous
  • anonymous
0
zepdrix
  • 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'
zepdrix
  • zepdrix
Yes, good 0* blah that disappeared somehow lol
anonymous
  • anonymous
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
zepdrix
  • zepdrix
Understand how to do that? It's not too bad, it just requires a bit of nasty algebra :)
anonymous
  • anonymous
yeah give me a sec i love algebra = )
zepdrix
  • zepdrix
haha nice XD
zepdrix
  • zepdrix
Oh my goodness betty... -_- are you using the equation tool? zzzzzzz lol
anonymous
  • anonymous
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 \[xy'e ^{y}+y'e ^{x}=-e ^{y}-e ^{x}y\] \[y'(xe ^{y}+e ^{x})=-e ^{y}-e ^{x}y\] \[y'=\frac{ -e ^{y}-e ^{x}y }{ xe ^{y}+e ^{x} }\]
anonymous
  • anonymous
haha yes i was
zepdrix
  • zepdrix
looks good c:
anonymous
  • anonymous
ok so now I plug 0 in for x?
anonymous
  • anonymous
what is that lol jk
zepdrix
  • zepdrix
|dw:1359347099057:dw|
zepdrix
  • zepdrix
This isn't necessarily what the function looks like. It's just to give us an idea of what they're asking.
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
They want us to form an equation for the `straight line` that is tangent to the function at x=0.
zepdrix
  • zepdrix
Remember the form of a straight line? :)
zepdrix
  • zepdrix
That's right! Your good ole buddy \(y=mx+b\). Remember him?
anonymous
  • anonymous
oh yeah ok i wasn't sure what u were asking lol
zepdrix
  • zepdrix
Our slope \(m\) is going to be equal to the slope of the tangent line at x=0. Meaning \(y'(0)=m\).
anonymous
  • anonymous
I was actually helping a person out today a lot with that form and the slope form with two points
zepdrix
  • zepdrix
Oh cool XD
zepdrix
  • 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.
anonymous
  • anonymous
oh ok i was wondering what i would do with the y-values
anonymous
  • anonymous
i got \[\frac{ -e-1 }{ 1 }\]
zepdrix
  • zepdrix
Yep looks good.
zepdrix
  • 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).
anonymous
  • anonymous
kk let me work it out = )
anonymous
  • anonymous
ok so this is what i got: \[y=-(e+1)x+1\] Now would you multiply it out like this: \[y= -ex-x+1\rightarrow -x(e+1)+1\]
zepdrix
  • zepdrix
No don't multiply it out c: looks good! Hopefully we didn't make any silly mistakes in there. Looks correct though :D
anonymous
  • anonymous
and then thats it? = )
zepdrix
  • zepdrix
Yes, we found an equation for the line tangent to our curve at x=0, y=1.
anonymous
  • anonymous
awesome = ) ty
anonymous
  • anonymous
I got a couple others = ) stay posted lol

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