Question

LGBADERIVATIVE:

\[\huge e^{x/y} = x - y\]

Answer 1

how do i derive the left side?

Answer 2

ln?

Answer 3

\[\huge \frac{x}{y} = \ln (x-y)?\]

Answer 4

i think that looks more complicated

Answer 5

Are you findind dy/dx or dx/dy ?

Answer 6

dy/dx

Answer 7

\[\frac{y - xy'}{y^2} = \frac{1 - y'}{x-y}\]

Answer 8

\[(x-y)(y-xy') = y^2(1-y')\]
\[xy - x^2y' -y^2 -xyy' = y^2 - y^2y'\]
\[-x^2y' - xyy' + y^2y' = y^2 - xy - y^2\]
\[y'(-x^2 - xy + y^2) = -xy\]
\[y' = \frac{xy}{x^2 +xy + y^2}\]
correct?

Answer 9

haha that looked nice...

Answer 10

i wish someone would tell me if im right or not....

Answer 11

@Zarkon ?

Answer 12

I don't think so

Answer 13

why not?

Answer 14

it is not correct

Answer 15

awww

Answer 16

you should just differentiate it straight forward,
\[e^{x/y}=x-y\]
\[(x(-1)y^{-2}\frac{dy}{dx}+y^{-1})e^{x/y}=1-\frac{dy}{dx}\]
\[(-\frac{x}{y^2}e^{x/y}+1)\frac{dy}{dx}=1-\frac{e^{x/y}}{y}\]
\[\huge \frac{dy}{dx}=\frac{1-\frac{e^{x/y}}y{}}{1-\frac{x}{y^2}e^{x/y}}\]
\[=\frac{y(y-e^{x/y})}{y^2-xe^{x/y}}\]

Answer 17

oh lol of course *facepalm* i feel so dumb hahahaha

Answer 18

@lgbasallote

you were not that far off.

Answer 19

but it should also be possible to use the method i did right o.O

Answer 20

yes...you just made an algebra mistake

Answer 21

ohhh