Anyone wanna solve an Implicit Differentiation problem?

[SkydiveNFlorida 0]

I know i'm just not seeing something, here.

Implicitly differentiate with respect to x.

e^xy + y = x - 1

I keep thinking to differentiate e^xy with respect to x you get ye^xy * x(dy/dx)*y so I get

ye^xy* [x(dy/dx)*y] + dy/dx = 1

But, the book says the next step was :
e^xy*[d/dx(xy) + dy/dx = 1

What am I not seeing here? ? :( I can't very well know if this is a soln to the diff eq I have if I can't even do implicit differentation, ahhh! Helllppp!

Angela.

[AggieDave 6]

42.

That's the answer. The answer to everything.

--"When I die, may I be surrounded by scattered chrome and burning gasoline."

[SkydiveNFlorida 0]

Quote

42.

That's the answer. The answer to everything.

LOL!!

[BGill 0]

Well i very well may be wrong but I think you're taking too many derivatives off the e^xy. The answer the book is giving is simply saying that the derivative of e^xy is e^xy times the derivative of the exponent (hence d/dx(xy)). The rest of the equation is just straight up derivative of the rest of the junk in the equation. Does that help? That's how I'm seeing this but my brain is also in "summer mode" right now so it may not be functioning at 100%... or even 50% for that matter

Good luck!

[Bodyflight.Net 0]

[closing door quietly on my way out]

[cocheese 0]

I got "fill it up"
That's why NASA said sorry.

[swilson 0]

When you take the derivative wrt x of the exponential, you're left with the y multiplied by the original exponential, the derivative of the y term goes to zero, the derivative of the x term equals one, and the constant term disappears.

d/dx(e^xy + y = x -1) works out to ye^xy = 1. Is this what you're after?

[SkydiveNFlorida 0]

Quote

Well i very well may be wrong but I think you're taking too many derivatives off the e^xy. The answer the book is giving is simply saying that the derivative of e^xy is e^xy times the derivative of the exponent (hence d/dx(xy)). The rest of the equation is just straight up derivative of the rest of the junk in the equation. Does that help? That's how I'm seeing this but my brain is also in "summer mode" right now so it may not be functioning at 100%... or even 50% for that matter

Good luck!

Yeah, but I guess my line of thought was that the derivative of e^ax would be ae^ax, so if you treated e^xy as y being a constant at first, you'd pull it out, then differentiate the exponent as a product. I guess I see what you're saying, I can't treat y as a constant to begin with. Very weird, so the derivative of e^xy with respect to x is only e^xy * x(dy/dx)+y not ye^xy * x(dy/dx) + y ?

It is so odd that even though you're differentiating with respect to variable x, you're treating y the same as dependent variable x at first (not pulling it out as a constant). Any insight that might help me accept this? I don't know why i'm having such trouble with this, most other problems are coming out fine.... except I HATE this TI-89 calculator!

thx!
Angela.

[SkydiveNFlorida 0]

Quote

When you take the derivative wrt x of the exponential, you're left with the y multiplied by the original exponential, the derivative of the y term goes to zero, the derivative of the x term equals one, and the constant term disappears.

d/dx(e^xy + y = x -1) works out to ye^xy = 1. Is this what you're after?

No, because that would be flat out differentation with respect to x. This is "implicit" differentation, so you have to implicitly differentiate y with respect to x. So, the problem :

x^2 + y^2 = 6 would implicitly differentiate with respect to x :

2x + 2y(dy/dx) = 0

I don't know this stuff very well either. I think I learned a little of this in Calc II, but I didn't see it in Calc III and I am just now seeing it again in Diff eq. I wish i'd have paid better attention b4.

Angela.

[swilson 0]

Okay, I misread this. I did some messing around in Maple (math software) and using the implicit differentiation command I came up with -(e^xy - 1)/(e^xy + 1). Is this closer?

[pleifer 0]

Quote

I know i'm just not seeing something, here.

Implicitly differentiate with respect to x.

e^xy + y = x - 1

I keep thinking to differentiate e^xy with respect to x you get ye^xy * x(dy/dx)*y so I get

ye^xy* [x(dy/dx)*y] + dy/dx = 1

But, the book says the next step was :
e^xy*[d/dx(xy) + dy/dx = 1

What am I not seeing here? ? :( I can't very well know if this is a soln to the diff eq I have if I can't even do implicit differentation, ahhh! Helllppp!

Angela.

e^xy*[d/dx(xy) + dy/dx = 1
I'll start by stating the rule that the derivative of e^x is e^x so that gives us the first term, then take the power ie the (xy)and take the derivative of that which is the d/dx(xy) + dy/dx (diff with respect to x the Y goes away) and then the = to 1 ( the derivative of x is 1 and the other 1 falles out
hope this helps.
supprizing i remember how to do this ha ha

_________________________________________
The Angel of Duh has spoke

[bch7773 0]

I passed that class, and i promised myself that i would never touch that crap again.

MB 3528, RB 1182

[wingnut 0]

sorry once you add that third variable in there i have no idea.... heck for allinow there are only two and i just don't know whatthe "e" is..... but as it is someday i hope to be as educated as youall and learn how to do this...

______________________________________
"i have no reader's digest version"

[weegegirl 2]

uhhh.... blue. no no... green.

[wingnut 0]

Quote

uhhh.... blue. no no... green

no no no.. it's red.. ya know the color of the paper after you're head starts bleeding after smacking it on the desk after working on a problem for like 4 hours and not getting the answer......

______________________________________
"i have no reader's digest version"

[Kris 0]

The answer is attached.

Sky, Muff Bro, Rodriguez Bro, and
Bastion of Purity and Innocence!™

[aero04 0]

Well here's my solution....

e^(xy)+y=x-1 -----original problem

ye^(xy)+xe^(xy)y'+y'=1-0 ----Implicit Diff of all terms

y'(xe^(xy)+1)=1-ye^(xy) -----factor out the y'

y'=(1-ye^(xy))/(xe^(xy)+1) ----solve for y'

edited to try to make it a bit more clear.

[champu 1]

btw, HH...

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