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\[xe ^{y}+ye ^{x}=1\]
at (0,1)

anybody??

any idea how to do this one? = )

Bahhhhh gimme few minutes XD eating some food.

kk take your time = )

Or confused about the directions? heh

hahaha ok so show me = )

so is it everytime the derivative of y will be 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

ok so the next part would be (and altogether):
\[(e ^{y}+xy'e ^{y})+(y'e ^{x}+ye ^{x})\]

=1 oops lol

Yes good c:
`Except` we took the derivative of `both sides`, so what should the right side produce?

Yes, good 0* blah that disappeared somehow 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 = )

haha nice XD

Oh my goodness betty... -_- are you using the equation tool? zzzzzzz lol

haha yes i was

looks good c:

ok so now I plug 0 in for x?

what is that lol jk

|dw:1359347099057:dw|

ok

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

Oh cool XD

oh ok i was wondering what i would do with the y-values

i got \[\frac{ -e-1 }{ 1 }\]

Yep looks good.

kk let me work it out = )

and then thats it? = )

Yes, we found an equation for the line tangent to our curve at x=0, y=1.

awesome = ) ty

I got a couple others = ) stay posted lol