Find dy/dx if x e^y + 1 = xy: I get y-e^y/xe^y-x

[venialove]

Find dy/dx if xe^y+1= xy
My work:
xe^y dy/dx + e^y+1 =0
dy/dx (xe^y +1) = e^y (dy/dx)
y-e^y/xe^y-x

 

[stapel]

Find dy/dx if xe^y+1= xy
My work: xe^y dy/dx + e^y+1 =0
dy/dx (xe^y +1) = e^y (dy/dx)
y-e^y/xe^y-x

Your last line (which is missing its "=dy/dx" for some reason...?) means something along the lines of the following:

. . . . .\(\displaystyle y\, -\, \frac{e^y}{x e^y}\, -\, x\, =\, y\, -\, \frac{1}{x}\, -\, x\)

Is this what you meant? Also, I'm afraid I don't understand your notation or reasoning for your various steps. When you reply, please include clarification of this as well. For instance, you'd had zero on the right-hand size, and you got a derivative of for that sides of e[sup:1u5qtte2]y[/sup:1u5qtte2] (dy/dx). How? And so forth. :shock:

Please be complete, starting with a clarification of what the original equation was. Thank you!

Eliz.

 

[venialove]

Find dy/dx

Find \frac{dy}{dx} if xe^y +1= xy

 

[stapel]

Find \frac{dy}{dx} if xe^y +1= xy

Okay. Is the equation either of the following?

. . . . .x e[sup:2dks8bay]y + 1[/sup:2dks8bay] = xy

. . . . .x e[sup:2dks8bay]y[/sup:2dks8bay] + 1 = xy

Also, how did you get from the original equation to your first line, quoted below?

My work: xe^y dy/dx + e^y+1 =0

Please reply with clarification, starting with the original equation, and providing a clear listing of your work and reasoning. Thank you!

Eliz.

 

[venialove]

it is this equation: x ey + 1 = xy

I got the other equation before when I tried to take the derivative of ey, but that part is incorrect. Also I thought I could get rid of xy and set it to zero but I remembered that it was not a constant.

 

[Deleted member 4993]

it is this equation: x ey + 1 = xy

I got the other equation before when I tried to take the derivative of ey, but that part is incorrect. Also I thought I could get rid of xy and set it to zero but I remembered that it was not a constant. That is correct - good thought - now show your work in line with that thought.

 

[galactus]

The solution you derived in your first post is correct.

 

[stapel]

The solution you derived in your first post is correct.

If, perhaps, some formatting is assumed...? Or does the derivative really simplify down to just "y"...?

Thank you!

Eliz.

 

[Deleted member 4993]

The solution is:

\(\displaystyle \frac{dy}{dx}\, = \, \frac{y \, - \, e^y}{x\cdot (e^y \, - \, 1)}\)

The answer in the first post was somewhat correct (except for grouping) - but the intermediate steps were wrong.