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

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0any idea how to do this one? = )

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Bahhhhh gimme few minutes XD eating some food.

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0kk take your time = )

zepdrix
 3 years ago
Best 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? :)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Or confused about the directions? heh

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0I 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: \[ye ^{x}=1xe ^{y}\] \[y=\frac{ 1xe ^{y} }{ e ^{x} }\] \[y'=\frac{ e ^{x}(y'xe ^{y})(1xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0ew 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:

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0hahaha ok so show me = )

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Implicit isn't too bad, just remember that whenever you take the derivative of a `y` term, a y' will pop out. Ok ok ok :)

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0so is it everytime the derivative of y will be y*y' ?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Everything 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
 3 years ago
Best 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,\[\large \color{salmon}{e^y+x(e^y)y'}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Understand how we did the first term? :)

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0ok that's right I remember now = )

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0that was for above hold on let me read what u posted lol

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Hehe yah I understand :3

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0On 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.

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

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

zepdrix
 3 years ago
Best 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'

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, good 0* blah that disappeared somehow lol

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0haha 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
 3 years ago
Best ResponseYou've already chosen the best response.0Understand how to do that? It's not too bad, it just requires a bit of nasty algebra :)

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0yeah give me a sec i love algebra = )

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Oh my goodness betty... _ are you using the equation tool? zzzzzzz lol

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0Ok I do love algebra but I could be wrong with this one but I think I might be right at the same time lol \[xy'e ^{y}+y'e ^{x}=e ^{y}e ^{x}y\] \[y'(xe ^{y}+e ^{x})=e ^{y}e ^{x}y\] \[y'=\frac{ e ^{y}e ^{x}y }{ xe ^{y}+e ^{x} }\]

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0ok so now I plug 0 in for x?

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0what is that lol jk

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0This isn't necessarily what the function looks like. It's just to give us an idea of what they're asking.

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Remember the form of a straight line? :)

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

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0oh yeah ok i wasn't sure what u were asking lol

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0Our slope \(m\) is going to be equal to the slope of the tangent line at x=0. Meaning \(y'(0)=m\).

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok i was wondering what i would do with the yvalues

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0i got \[\frac{ e1 }{ 1 }\]

zepdrix
 3 years ago
Best 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).

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0kk let me work it out = )

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0No don't multiply it out c: looks good! Hopefully we didn't make any silly mistakes in there. Looks correct though :D

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0and then thats it? = )

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

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0I got a couple others = ) stay posted lol
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