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