Question

I need help finding dy/dx for 1+ lnxy=e^(x-y)

Answer 1

Okay. Assumed knowledge in this should be that \[\frac{ dy }{ dx }\ln(x) = \frac{ 1 }{ x }\]There are plenty of formal proofs for this. By the way, a great website for this kind of stuff is this: http://www.derivative-calculator.net/# It gives you step-by-step explanations on how it was derived, and the rules involved. (Rest below, one sec.)

Answer 2

Thanks I will check it out

Answer 3

I understand everything just not the e part

Answer 4

So, looks like implicit differentiation may have to be used. Let me write this out:\[1 + \ln(xy) = e ^{(x - y)} \]From here, taking the derivative of everything: \[\frac{ dy }{ dx }(1) + \frac{ dy }{ dx }(xy) = \frac{ dy }{ dx }e ^{(x-y)} = (0) + \frac{ dy }{ dx }\ln(x)+ \frac{ dy }{ dx }\ln(y) = \frac{ dy }{ dx }e ^{(x-y)}\]Oh. One sec. I'll cut to the chase. And i'm not just saying that to be like, "Hey, check this stuff out. I really mean that that thing is REALLY useful. It's saved me prior to tests more than once.

Answer 5

So, \[\frac{ dy }{ dx }e ^{x} = e ^{x}\]Agreed?

Answer 6

Agreed!

Answer 7

Okay. So \[\frac{ dy }{ dx }e ^{(x-y)} = e ^{(x-y)}*(1- \frac{ dy }{ dx })\]From here, dy/dx can be treated as a variable. So, just doing distribution:\[e ^{(x-y)}- e ^{(x-y)}\frac{ dy }{ dx } = \frac{ 1 }{ x } + \frac{ 1 }{ y }\frac{ dy }{ dx }\]Give me a moment, i'm getting nervous with my own crappy Calculus skills and want to make sure that I don't slip up, lol.

Answer 8

I'm not sure if I truly understand... And ok haha

Answer 9

Oh I see what u did!

Answer 10

Nevermind I've lost it lol

Answer 11

....YYyyyyyeaahhhhh, i'm not comfortable with this yet, either. I'm trying to find a calculator to check if i'm right, but i'm getting confused by their own answers. Sorry. I'll have an answer, but it might take me a little, lol.

Answer 12

That's alright! Thankyou! I will keep trying as well. I've gotten 1/xy (dy/dx) = e^(x-y) • (1-dy/dx)

Answer 13

Help her, @strawberry17 ! You're her only hope.

*throws in proverbial towel for the time being*

Answer 14

lol!

Answer 15

Hahahaha thanks!

Answer 16

if this is calculus im not up there yet :/

Answer 17

Aw jeez haha it's calculus...

Answer 18

sorry :(

Answer 19

That's alright thanks anyways!

Answer 20

Wait, I think I got it. Well, I don't really think I didn't get it in the first place, but I didn't have the confidence to say that I got it, lol. Give me like 5 minutes.

As a general rule,
\[\frac{ dy }{ dx }\ln(u) = \frac{ 1 }{ u }\frac{ du }{ dx }\]
I separated the logs for the sake of simplicity.
\[\frac{ d }{ dx }\ln(xy) = \frac{ d }{ dx }[\ln(x) + \ln(y)] = \frac{ 1 }{ x }+\frac{ 1 }{ y }\frac{ dy }{ dx }\]
Same as before, the current equation can be expressed as:\[\frac{ 1 }{ x }\frac{ dy }{ dx } = e ^{(x-y)} - e ^{(x-y)}\frac{ dy }{ dx }\]

Answer 21

From here, all you need to do is isolate dy/dx.

\[\frac{ 1 }{ x }\frac{ dy }{ dx } = e ^{(x-y)} - e ^{(x-y)}\frac{ dy }{ dx } ||| ||||||\frac{ 1 }{ x }\frac{ dy }{ dx }+e ^{(x-y)} \frac{ dy }{ dx } = e ^{(x-y)}\]
\[\frac{ dy }{ dx }[\frac{ 1 }{ x } + e ^{(x-y)}] = e ^{(x-y)} ||||||||| \frac{ dy }{ dx } (ughgod, this-is-gonna-look-ugly.)\]

Answer 22

Thanks!

Answer 23

\[\frac{ dy }{ dx } = \frac{ e ^{(x-y)} }{ [\frac{ 1 }{ x }+e ^{(x-y)}] }\]

Answer 24

Yeah, no problem, lol. That's a pretty horrible answer, though. I'm not going to even attempt to simplify it, and i'm not sure if it's right.

Answer 25

That's alright I will just go ask my professor if it's right :) thanks!

Answer 26

@Hero Could you check if this is right? I don't mean to be a bother. Yeah, np. Good luck.

Answer 27

Thanks! Ima need it...