Question

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 1

yes you do. but it is a pain to make it so

Answer 2

what kind of a question is "do you get the same result"???!!!

Answer 3

by doing (i) and (ii). do you get the same answer?

Answer 4

first of all i assume you are looking for
\[y'\] yes?

Answer 5

ya

Answer 6

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

Answer 7

gotta solve this for
\[y'\] yikes. i need pencil and paper

Answer 8

you probably dont get the same result both times

Answer 9

because the first way of writing it has a denominator, so the points where denominator equals zero should be excluded

Answer 10

the second expression doesnt have a division.

Answer 11

well this is what i got the first time @elecengineer please check:
\[2x(x-y)^2=2xy'-2yy'\]
\[2x(x-y)^2=y'(2x-2y)\]
\[y'=\frac{2x(x-y)^2}{2x-2y}=\frac{x(x-y)^2}{x-y}=x(x-y)\]

Answer 12

after a bunch o'algebra

Answer 13

i don't have to worry about division by 0 because in the original expression you cannot have
\[x=y\]

Answer 14

let me try second way.

Answer 15

yup.. ==" calm down.. look slowly..

Answer 16

@elecengineer it is the first version that has not denominator. second one does. now i have to figure out if i made a mistake, or why they are the same. can i cancel?

Answer 17

oooooooooooooooooops

Answer 18

\[x^3-x^2y=x+y\] maybe this works better

Answer 19

\[3x^2 -2xy-x^2y'=1+y'\]
\[3x^2-2xy-1=y'+x^2y'=y'(1+x^2)\]
\[y'=\frac{3x^2-2xy-1}{1+x^2}\]

Answer 20

better algebra this time?

Answer 21

Digesting it.......

Answer 22

well what is going to be really annoying is figuring out why they are the same.

Answer 23

oh lord i think i made a mistake in the first one. let me try it again

Answer 24

i thought it is annoyting to think that why the answer are not the same?

Answer 25

let me check the first solution

Answer 26

okay.

Answer 27

ok i messed up and wrote something wrong on my paper.

Answer 28

yup.. i think if i wasn't mistaken.. the first equation you had wrong druring the 3rd line..

Answer 29

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

Answer 30

somehow i screwed up expanding on the right hand side. it is
\[x+xy'-y-yy'-(x-xy'+y-yy')\]
\[x+xy'-y-yy'-x+xy'-y+yy'\]
\[=2xy'-2y\]

Answer 31

yup! i got that too!

Answer 32

so now we have
\[2x(x-y)^2=2xy'-2y\]
\[2x(x-y)^2+2y=2xy'\]
\[\frac{2x(x-y)^2+2y}{2x}=y'\]
\[y'=\frac{x(x-y)^2+y}{x}\] unless i made another mistake

Answer 33

i think you got it right.. but now i was wondering.. why isn't the answer is same.. LOL

Answer 34

and i have no idea why they are the same.
oh they are the same

Answer 35

you just can't see it. don't forget that you have an equality to begin with . it will depending on using it at some point

Answer 36

you started with
\[x^2=\frac{x+y}{x-y}\] so somehow using that you can show these are the same

Answer 37

Ermmmmmm.. digesting...

Answer 38

i guess what i am saying is that if you look in one expression for the derivative you see
\[x^2\] so you are free to replace is by
\[\frac{x+y}{x-y}\] if you choose. there is probably some clever algebra to be done, but i don't see it

Answer 39

Thanks!

Answer 40

yw. sorry about the algebra. if i had an hour i bet i could do it....maybe

Answer 41

hahaha.. its okay.. you already gave me a big help.

Answer 42

Can you do me 1 more favour? Can you show me the way to verify both (i) & (ii) are the same?

Answer 43

@satellite73 hlp!!><

Answer 44

x^2 =(x+y)/(x−y)

find dy/dx in two ways:
(i) by multiplying (x-y) on both sides

x^2 (x-y) = x+y

Dx(x^3-x^2 y = x+y)

3x^2 -x^2 y' -2xy = 1+y'

3x^2 -2xy -1 = y' +x^2 y'

3x^2 -2xy -1 = y'(1 +x^2)

y' = (3x^2 -2xy -1)/(1+x^2)

(ii) by quotient rule

x^2 =(x+y)/(x−y)

Dx [x^2 =(x+y)/(x−y)]

2x = [(x-y)(1+y') - (x+y)(1-y')]/[x-y]^2

2x(x-y)^2 = x+xy' -y -yy' -x +xy' -y +yy'

2x(x-y)^2 = xy' -2y +xy'

2x(x-y)^2 +2y = y'(2x)

y' = (x-y)^2 + y/x
......................................
(3x^2 -2xy -1)/(1+x^2) = (x-y)^2 +yx ; is this what we got so far?

Answer 45

yaya!

Answer 46

well, they dont make the same graph when plotted ....

Answer 47

and then... =.=" stucked... this question is wrong??

Answer 48

nothing went wrong

Answer 49

they are surely the same. but to do that you are going to have to do a raft of algebra. i bet amistre can do it yes?

Answer 50

Totally confused... o.O"

Answer 51

i cant see any errors .. but i cant see the equality either :)

Answer 52

no no you did nothing wrong

Answer 53

i made a couple errors but fixed them and i got what you got

Answer 54

not without buying a few notebooks maybe :)

Answer 55

http://www.wolframalpha.com/input/?i=y+%3D+%28x%2By%29%2F%28x-y%29-x^2

Answer 56

this doesn't help me.

Answer 57

we have two expressions for y ' yes?

Answer 58

yes ....

Answer 59

I even tried plugging in the same x and y value into each of them .... to test

Answer 60

\[y'=\frac{3x^2 -2xy -1}{1+x^2}\] and
\[y'=(x-y)^2+\frac{y}{x}=\frac{x(x-y)^2}{x}\]

Answer 61

now i am next to positive they are the same. not that they are the same expression, because they certainly are not

Answer 62

but don't forget we started with
\[x^2=\frac{x+y}{x-y}\] so using this identity and some algebra that i don't know, i am willing to bet you can show these are the same

Answer 63

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

Answer 64

that is what i got yes

Answer 65

oh good; then you musta dropped a y in transit :)

Answer 66

oh sorry yes a "typo"

Answer 67

so for example in the expression
\[1+x^2\] in the denominator of first example, you can replace \[x^2\] by \[\frac{x+y}{x-y}\] if you like

Answer 68

let me try to do some algebra. it will be a pain for sure but i cannot do it here. somehow you can show that these are the same, given the original identity

Answer 69

Brain storm had turn to be a storm in the brain... =."

Answer 70

i must be really messing up because i got the derivative is 1.

Answer 71

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

Answer 72

everyone should give you medals, your first post was 2 hours ago o.O

Answer 73

and
\[x^2=\frac{x+y}{x-y}\rightarrow x^2(x-y)=x+y \rightarrow x-y=\frac{x+y}{x^2} \rightarrow \]
\[(x-y)^2=\frac{(x+y)(x-y)}{x^2}=\frac{x^2-y^2}{x^2}\]
\[y'=\frac{x^2-y^2}{x^2}+\frac{y}{x}=\frac{x^2-y^2+xy}{x}\]

Answer 74

ok joemath i have to go. your assignment, if you decide to accept it, so to show that these derivatives are the same:
\[y'=\frac{x(x-y)^2+y}{x}\] and
\[\frac{3x^2-2xy-1}{x^2+1}\]

Answer 75

you are going to need that
\[x^2=\frac{x+y}{x-y}\]

Answer 76

good luck. as always, should you or any of your IM force be caught or killed, the sensei will disavow any knowledge of your actions.

Answer 77

if you do this in two minutes i am never coming back to this site.

Answer 78

lolol, im at school right now, i'll work on this during my american lit class

Answer 79

have fun

Answer 80

i got the same thing as you using the quotient rule

Answer 81

solve for y in the equation

\[x^2=\frac{x+y}{x-y}\]

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

now substitute this into both derivatives and simplify...you will see that they are the same

Answer 82

read and weep