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Shane Strate
science forum beginner

Joined: 14 Jun 2006
Posts: 2

Posted: Wed Jun 14, 2006 6:08 pm    Post subject: Question about Implicit Differentiation

What does everybody think about differentiating

xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0

Thanks alot for the help,
S. Strate
David Moran
science forum Guru Wannabe

Joined: 13 May 2005
Posts: 252

Posted: Wed Jun 14, 2006 6:56 pm    Post subject: Re: Question about Implicit Differentiation

"Shane Strate" <sstrate@gmu.edu> wrote in message
 Quote: What does everybody think about differentiating xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0 Thanks alot for the help, S. Strate

Are you finding dy/dx or dx/dy?

Dave
Rich Carreiro
science forum beginner

Joined: 03 May 2005
Posts: 28

Posted: Wed Jun 14, 2006 6:57 pm    Post subject: Re: Question about Implicit Differentiation

"Shane Strate" <sstrate@gmu.edu> writes:

 Quote: What does everybody think about differentiating xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0

You didn't do the implicit differentiation. Where are
the y' (or dy/dx)?

For example, implicitly differentiating xe^y
gives you (using y' to stand for dy/dx):
x y' e^y + e^y = (xy' + 1)e^y

--
Rich Carreiro rlcarr@animato.arlington.ma.us
Shane Strate
science forum beginner

Joined: 14 Jun 2006
Posts: 2

Posted: Wed Jun 14, 2006 9:56 pm    Post subject: Re: Question about Implicit Differentiation

"Rich Carreiro" <rlcarr@animato.arlington.ma.us> wrote in message
news:m3mzcfk1jd.fsf@animato.home.lan...
 Quote: "Shane Strate" writes: What does everybody think about differentiating xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0 You didn't do the implicit differentiation. Where are the y' (or dy/dx)? For example, implicitly differentiating xe^y gives you (using y' to stand for dy/dx): x y' e^y + e^y = (xy' + 1)e^y -- Rich Carreiro rlcarr@animato.arlington.ma.us

Sorry about that. I meant to type

(-e^x + ye^x)/(xe^y+e^xy)=y'

but your example was very helpful, I appreciate it. I was differentiating
xe^y as e^yxyy'. My logic was as follows.

Using the product rule,
u = x
v = e^y
du = 1
dv = e^y(y)y' <----- I take it that the (y) term here is unnecessary?

udv + vdu = xe^y(y)y' + e^y =>
y' = -e^y/xe^y(y) =>
y' = -1/xy (((as opposed to what you are saying))) y' = -1/x

Once again, thanks alot, I really do appreciate it.
Stan Brown
science forum Guru Wannabe

Joined: 06 May 2005
Posts: 279

Posted: Thu Jun 15, 2006 1:36 am    Post subject: Re: Question about Implicit Differentiation

Wed, 14 Jun 2006 14:08:00 -0400 from Shane Strate <sstrate@gmu.edu>:
 Quote: What does everybody think about differentiating xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0

Unless I'm missing something, it's not correct. The left-hand side
does indeed look like y', but you haven't told us anything to justify
setting it equal to 0.

 Quote: Thanks alot for the help,

"A lot" is two words. There's really no reason not to get it right.

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
Paul Sperry
science forum Guru

Joined: 08 May 2005
Posts: 371

Posted: Thu Jun 15, 2006 3:07 am    Post subject: Re: Question about Implicit Differentiation

In article <ZfYjg.18\$PO.4@dukeread03>, Shane Strate <sstrate@gmu.edu>
wrote:

 Quote: What does everybody think about differentiating xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0 Thanks alot for the help, S. Strate

A piece at a time:

d/dx(x*e^y) = x*d/dx(e^y) + d/dx(x)*e^y = x*e^y*y' + e^y;

d/dx(y*e^x) = y*d/dx(e^x) + d/dx(y)*e^x = y*e^x + y'*e^x;

d/dx(2) = 0.

Plug in and solve for y'.

--
Paul Sperry
Columbia, SC (USA)
G.E. Ivey
science forum Guru

Joined: 29 Apr 2005
Posts: 308

Posted: Fri Jun 16, 2006 1:09 pm    Post subject: Re: Question about Implicit Differentiation

 Quote: "Rich Carreiro" writes: What does everybody think about differentiating xe^y - ye^x=2 as (-e^x + ye^x)/(xe^y+e^xy)=0 You didn't do the implicit differentiation. Where are the y' (or dy/dx)? For example, implicitly differentiating xe^y gives you (using y' to stand for dy/dx): x y' e^y + e^y = (xy' + 1)e^y -- Rich Carreiro rlcarr@animato.arlington.ma.us Sorry about that. I meant to type (-e^x + ye^x)/(xe^y+e^xy)=y' but your example was very helpful, I appreciate it. I was differentiating xe^y as e^yxyy'. My logic was as follows. Using the product rule, u = x v = e^y du = 1 dv = e^y(y)y' <----- I take it that the (y) term here is unnecessary? Well, I wouldn't say "unnecessary"- I would say "wrong"!

By the chain rule, the derivative of e^y with respect to x is the derivative of e^y with respect to y, which is just e^y, times the derivative of y with respect to x: y'. The derivative of e^y with respect to x is
e^y (y')
 Quote: udv + vdu = xe^y(y)y' + e^y = Assuming you are still talking about xe^y= constant, y' = -e^y/xe^y(y) = y' = -1/xy (((as opposed to what you are saying))) y' = -1/x Once again, thanks alot, I really do appreciate it. Assuming you are talking about xe^y= constant, then

(xe^y)'= e^y+ xe^y(y')= 0 so y'= -e^x/(xe^x)= -1/x.

Of course, with your original xe^y- ye^x= 2 it's more complicated:
(xe^y- ye^x)'= (e^y+ xe^y(y')- y'e^x- ye^x)= 0
(xe^x- e^x)y'+ (e^y- ye^x)= 0
(xe^x- e^x)y'= ye^x- e^y
y'= (ye^x- e^y)/(xe^x- e^x)

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