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 = )

- anonymous

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- anonymous

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

- anonymous

anybody??

- anonymous

@zepdrix

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## More answers

- anonymous

any idea how to do this one? = )

- zepdrix

Bahhhhh gimme few minutes XD eating some food.

- anonymous

kk take your time = )

- 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

Or confused about the directions? heh

- 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

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

hahaha ok so show me = )

- 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

so is it everytime the derivative of y will be y*y' ?

- 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

\[\huge \color{salmon}{xe^y}+ye^x=1\]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

Understand how we did the first term? :)

- anonymous

ok that's right I remember now = )

- anonymous

that was for above hold on let me read what u posted lol

- zepdrix

Hehe yah I understand :3

- 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

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

- anonymous

=1 oops lol

- zepdrix

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

- anonymous

0

- 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

Yes, good 0* blah that disappeared somehow lol

- 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

Understand how to do that?
It's not too bad, it just requires a bit of nasty algebra :)

- anonymous

yeah give me a sec i love algebra = )

- zepdrix

haha nice XD

- zepdrix

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

- 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

haha yes i was

- zepdrix

looks good c:

- anonymous

ok so now I plug 0 in for x?

- anonymous

what is that lol jk

- zepdrix

|dw:1359347099057:dw|

- zepdrix

This isn't necessarily what the function looks like. It's just to give us an idea of what they're asking.

- anonymous

ok

- zepdrix

They want us to form an equation for the `straight line` that is tangent to the function at x=0.

- zepdrix

Remember the form of a straight line? :)

- zepdrix

That's right! Your good ole buddy \(y=mx+b\).
Remember him?

- anonymous

oh yeah ok i wasn't sure what u were asking lol

- zepdrix

Our slope \(m\) is going to be equal to the slope of the tangent line at x=0.
Meaning \(y'(0)=m\).

- anonymous

I was actually helping a person out today a lot with that form and the slope form with two points

- zepdrix

Oh cool XD

- 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

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

- anonymous

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

- zepdrix

Yep looks good.

- 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

kk let me work it out = )

- 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

No don't multiply it out c:
looks good!
Hopefully we didn't make any silly mistakes in there. Looks correct though :D

- anonymous

and then thats it? = )

- zepdrix

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

- anonymous

awesome = ) ty

- anonymous

I got a couple others = ) stay posted lol

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