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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
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

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bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
\[xe ^{y}+ye ^{x}=1\] at (0,1)
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
anybody??
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
any idea how to do this one? = )
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Bahhhhh gimme few minutes XD eating some food.
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
kk take your time = )
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\huge xe^y+ye^x=1\]Were you able to take a derivative of this function? Or are we having some trouble on that part? :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Or confused about the directions? heh
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
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: \[ye ^{x}=1xe ^{y}\] \[y=\frac{ 1xe ^{y} }{ e ^{x} }\] \[y'=\frac{ e ^{x}(y'xe ^{y})(1xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
hahaha ok so show me = )
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
so is it everytime the derivative of y will be y*y' ?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\huge \color{salmon}{xe^y}+ye^x=1\]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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Understand how we did the first term? :)
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
ok that's right I remember now = )
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
that was for above hold on let me read what u posted lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Hehe yah I understand :3
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
ok so the next part would be (and altogether): \[(e ^{y}+xy'e ^{y})+(y'e ^{x}+ye ^{x})\]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
=1 oops lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yes good c: `Except` we took the derivative of `both sides`, so what should the right side produce?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\huge e^y+xe ^y\color{orangered}{y'}+e^x\color{orangered}{y'}+ye^x=0\]Our goal from here is to solve for y'
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yes, good 0* blah that disappeared somehow lol
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Understand how to do that? It's not too bad, it just requires a bit of nasty algebra :)
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
yeah give me a sec i love algebra = )
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
haha nice XD
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh my goodness betty... _ are you using the equation tool? zzzzzzz lol
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
haha yes i was
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
looks good c:
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
ok so now I plug 0 in for x?
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
what is that lol jk
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
dw:1359347099057:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
They want us to form an equation for the `straight line` that is tangent to the function at x=0.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Remember the form of a straight line? :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
That's right! Your good ole buddy \(y=mx+b\). Remember him?
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
oh yeah ok i wasn't sure what u were asking lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
I was actually helping a person out today a lot with that form and the slope form with two points
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh cool XD
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
oh ok i was wondering what i would do with the yvalues
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
i got \[\frac{ e1 }{ 1 }\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yep looks good.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large y=mx+b \qquad \rightarrow \qquad y=(e+1)x+b\] And now we only need to find the yintercept `b`. To do so, we'll again use the point that ws given (0,1).
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
kk let me work it out = )
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
ok so this is what i got: \[y=(e+1)x+1\] Now would you multiply it out like this: \[y= exx+1\rightarrow x(e+1)+1\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
and then thats it? = )
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yes, we found an equation for the line tangent to our curve at x=0, y=1.
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
awesome = ) ty
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
I got a couple others = ) stay posted lol
 one year ago
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