Question

the x^2=(x+y)/(x-y)
making a new thread because other one is getting long

Answer 1

for people who want to look at it
the problem was

x^2 =(x+y)÷(x−y)
find dy/dx in two ways:
(i) by multiplying (x-y) on both sides
(ii) by quotient rule
Do you get the same result each way? please give me explanation if it isn't same and how to verify if both ways is the same.

Answer 2

and the answer to tys0702 or whatever your name was

Answer 3

ho ho ho ho ho ho ho ho ho

Answer 4

oops is yes lol

Answer 5

i'd love to see the proof that they are the same. i think i even remember by heart what you get in either case because i only spent maybe 2 hours trying to reconcile them

Answer 6

look at my attachment then

Answer 7

satellite the trick was not was
not to do x^2(x-y)=x^3-x^2y
but to leave it as x^2(x-y) and then take derivative

Answer 8

using quotient rule you get
\[y'=\frac{x(x-y)^2+y}{x}\]

Answer 9

right

Answer 10

and other way you get
\[\frac{3x^2-2xy-1}{1+x^2}\]

Answer 11

and for some reason these are the same. i am sure of it and i am sure the algebra had me stumped for an hour or more

Answer 12

1st are we assuming (x-y) cannot equal zero?
does the orginal problem state multiply by (x-y) or was that you assumption?

Answer 13

\[(x-y)*x^2=\frac{x+y}{x-y}*(x-y)\]
\[(1-y')x^2+2x(x-y)=1+y'\]
\[2x(x-y)+x^2-x^2y'=1+y'\]
\[2x(x-y)+x^2-1=y'+x^2y'\]
\[2x(x-y)+x^2-1=y'(1+x^2)\]
\[\frac{2x(x-y)+x^2-1}{x^2+1}=y'\]
\[\frac{2x(x-y)+\frac{x+y}{x-y}-1}{\frac{x+y}{x-y}+1}=y'\]
\[y'=\frac{2x(x-y)^2+(x+y)-(x-y)}{x+y+x-y)}=\frac{2x(x-y)^2+2x}{2x}\]
\[y'=\frac{x(x-y)^2+x}{x}\]

Answer 14

ok my myininaya i tried to read what you wrote as an attachment and i am going to have to assume that you out algebra'd me because i tried just that. i tried replacing
\[x^2\] by
\[\frac{x+y}{x-y}\] and ended up with a mess.

Answer 15

lol

Answer 16

you win the algebra contest for sure. probably involved "factor by grouping"

Answer 17

i tried to distribute as little as possible to begin with

Answer 18

probably best idea. ok you get it. i retire my algebra gloves. damn medal button is still not working. tried both firefox and chrome, so you get a virtual medal from me

Answer 19

:(

Answer 20

yes i assumed x did not equal y, duke

Answer 21

of course x does not equal y because it cannot equal y to begin with

Answer 22

oooh i didn't see that nice latex before! i am going to frame it.