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Shane Strate
science forum beginner
Joined: 14 Jun 2006
Posts: 2
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 Posted: Wed Jun 14, 2006 6:08 pm Post subject:
Question about Implicit Differentiation
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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 |
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David Moran
science forum Guru Wannabe
Joined: 13 May 2005
Posts: 252
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| Posted: Wed Jun 14, 2006 6:56 pm Post subject:
Re: Question about Implicit Differentiation
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"Shane Strate" <sstrate@gmu.edu> wrote in message
news:ZfYjg.18$PO.4@dukeread03...
| 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
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Are you finding dy/dx or dx/dy?
Dave |
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Rich Carreiro
science forum beginner
Joined: 03 May 2005
Posts: 28
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| Posted: Wed Jun 14, 2006 6:57 pm Post subject:
Re: Question about Implicit Differentiation
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"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
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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 |
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Shane Strate
science forum beginner
Joined: 14 Jun 2006
Posts: 2
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| Posted: Wed Jun 14, 2006 9:56 pm Post subject:
Re: Question about Implicit Differentiation
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"Rich Carreiro" <rlcarr@animato.arlington.ma.us> wrote in message
news:m3mzcfk1jd.fsf@animato.home.lan...
| Quote: | "Shane Strate" <sstrate@gmu.edu> 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. |
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Stan Brown
science forum Guru Wannabe
Joined: 06 May 2005
Posts: 279
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| Posted: Thu Jun 15, 2006 1:36 am Post subject:
Re: Question about Implicit Differentiation
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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
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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
http://OakRoadSystems.com/ |
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Paul Sperry
science forum Guru
Joined: 08 May 2005
Posts: 371
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| Posted: Thu Jun 15, 2006 3:07 am Post subject:
Re: Question about Implicit Differentiation
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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) |
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G.E. Ivey
science forum Guru
Joined: 29 Apr 2005
Posts: 308
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| Posted: Fri Jun 16, 2006 1:09 pm Post subject:
Re: Question about Implicit Differentiation
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| Quote: | "Rich Carreiro" <rlcarr@animato.arlington.ma.us
wrote in message
news:m3mzcfk1jd.fsf@animato.home.lan...
"Shane Strate" <sstrate@gmu.edu> 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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