Question

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 = )

Answer 1

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

Answer 2

anybody??

Answer 3

@zepdrix

Answer 4

any idea how to do this one? = )

Answer 5

Bahhhhh gimme few minutes XD eating some food.

Answer 6

kk take your time = )

Answer 7

\[\huge xe^y+ye^x=1\]Were you able to take a derivative of this function? Or are we having some trouble on that part? :)

Answer 8

Or confused about the directions? heh

Answer 9

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:
\[ye ^{x}=1-xe ^{y}\]
\[y=\frac{ 1-xe ^{y} }{ e ^{x} }\]
\[y'=\frac{ e ^{x}(y'xe ^{y})-(1-xe ^{y})(e ^{x}) }{ (e ^{x})^{2} }\]

Answer 10

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:

Answer 11

hahaha ok so show me = )

Answer 12

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 :)

Answer 13

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

Answer 14

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'.

Answer 15

\[\huge \color{salmon}{xe^y}+ye^x=1\]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,\[\large \color{salmon}{e^y+x(e^y)y'}\]

Answer 16

Understand how we did the first term? :)

Answer 17

ok that's right I remember now = )

Answer 18

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

Answer 19

Hehe yah I understand :3

Answer 20

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.

Answer 21

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

Answer 22

=1 oops lol

Answer 23

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

Answer 24

0

Answer 25

\[\huge e^y+xe ^y\color{orangered}{y'}+e^x\color{orangered}{y'}+ye^x=0\]Our goal from here is to solve for y'

Answer 26

Yes, good 0* blah that disappeared somehow lol

Answer 27

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

Answer 28

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

Answer 29

yeah give me a sec i love algebra = )

Answer 30

haha nice XD

Answer 31

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

Answer 32

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
\[xy'e ^{y}+y'e ^{x}=-e ^{y}-e ^{x}y\]
\[y'(xe ^{y}+e ^{x})=-e ^{y}-e ^{x}y\]
\[y'=\frac{ -e ^{y}-e ^{x}y }{ xe ^{y}+e ^{x} }\]

Answer 33

haha yes i was

Answer 34

looks good c:

Answer 35

ok so now I plug 0 in for x?

Answer 36

what is that lol jk