Question

Use implicit differentiation to determine dy/dx if
x*e^y+1=x*y

Answer 1

OK. Do you know how implicit works?

Answer 2

A bit, yes. I know it's basically chain ruling.

Answer 3

Yah. The whole goal is that wherever you have a y it becomes a dy/dx, then you solve for that. Now, this isa little odd in the question part so... Are you trying to determine if the left of the = is the same as the right?

Answer 4

OR, it set as x*e^y+1=x*y and you want to do the whole thing. I have seen these things go both ways in questions.

Answer 5

From the way you wrote it, I am betting the second... which will be fun because it means you have to differentiate both sides of the equation.

Answer 6

Yeah, both sides, lol.

Answer 7

OK, so know the setup? You are looking at product rules on both sides with a chained on dy every time you differentiate a y.

Answer 8

So it would dy/dx(x*e^y)+dy/dx[1]=dy/dx[x*y]

Answer 9

e^y*x+0=xy'

Answer 10

Yeah, I know I used two different notations, lol. y' is easier to type.

Answer 11

np

Answer 12

The right side would be xy, so product rule on differentiation.

Answer 13

Actually, product rule on both sides.

Answer 14

\[\frac{d}{dx}[xe^y+1]=\frac{d}{dx}[xy]\Rightarrow \frac{d}{dx}[xe^y]+ \frac{d}{dx}[1]=\frac{d}{dx}[xy]\Rightarrow
\frac{yd}{dx}[xe^y]+1e^y + 0=\frac{yd}{dx}x+1y\]

Answer 15

Oops. Meant dy/dx, not yd. LOL!

Answer 16

Lol, np. x3

Answer 17

So this, before it is solved for \(\frac{dy}{dx}\) \[\frac{dy}{dx}[xe^y]+e^y =\frac{dy}{dx}x+1y\]

Answer 18

Now, there are two solutions from that point, but they differ very little. It depends on which side you pull the \(\frac{dy}{dx}\) to.

Answer 19

Do you get what I am taling about there?

Answer 20

Yes.

Answer 21

OK, so where would you go with it from that point?

Answer 22

all the dy/dx will be on th eleft side.

Answer 23

OK. So show me your version. I'm doing both. Hehe.

Answer 24

Lol.

Answer 25

dy/dx = -e^y+y/(xe^y-x) ? o:

Answer 26

Any luck over there?

Answer 27

I just typed out an answer.

Answer 28

Hehe. About the same time. LOL.
\[e^y-1y =\frac{dy}{dx}x-\frac{dy}{dx}xe^y\\
e^y-1y =\frac{dy}{dx}(x-xe^y)\\
\frac{e^y-1y}{x-xe^y} =\frac{dy}{dx}\]
\[\frac{dy}{dx}xe^y-\frac{dy}{dx}x =1y-e^y\\
\frac{dy}{dx}(xe^y-x) =1y-e^y\\
\frac{dy}{dx} =\frac{1y-e^y}{xe^y-x}\]

Answer 29

Haha, nice. :D

Answer 30

Hehe. Thought you might like to see the way I did it. I mean, it is just algebra, but you really have to watch those sign changes!
90% of calculus is algebra, trig, and everything down to 1+1=2.

Answer 31

In derivatives, as you know, you do things like \(nx^{n-1}\). When you get to anti-derivatives (integrals) you will do this in reverse. If you think about it, \(n\pm 1\) can be \(2-1\) and \(1+1\), AKA: basic arithmetic.

So back in first grade, you were learning skills required to do calculus and they never bothered to tell you!