Question

Implicit differentiation: y^2 = (x^2)/(xy-4). Thank you!

Answer 1

ick

Answer 2

\[y^2=\frac{x^2}{xy-4}\] right?

Answer 3

yes! ick is right haha

Answer 4

the best way to think of this is as
\[f^2(x)=\frac{x^2}{xf(x)-4}\] you need the product rule, the quotient rule, and the chain rule

Answer 5

the left hand sides is easy, the derivative of \(f^2(x)\) is \(2f(x)f'(x)\) which you write as
\[2yy'\]

Answer 6

the right hand side is a bear

Answer 7

yeah, the right part is the... .. T__T

Answer 8

love your sarcasm, @satellite73

Answer 9

ok but it is still doable
the derivative of the numerators is \(2x\) and the derivative of the denominator is
\[y+xy'\] by the product rule
@nincompoop i'll take that as a compliment thank

Answer 10

laughing out loud

Answer 11

is it fine if it change it to: \[(y ^{2})(xy-4)-x^2=0\]?

Answer 12

you get via the quotient rule
\[\frac{(xy-4)\times 2x-x^2(y+xy')}{(xy-4)^2}\]

Answer 13

then comes a raft of algebra
i guess that might help we can try it both ways

Answer 14

actually that might be a good idea
we get \[(y ^{2})(xy-4)-x^2=0\] or
\[xy^3-4y^2-x^2=0\] lets see
\[y^3-3xy^2y'-9yy'=2x=0\] hmm looks promising

Answer 15

\[y^3-3xy^2y'-8yy'-2x=0\]

Answer 16

then
\[y^3-2x=3xy^2y'-8yy'\]
\[y^3-2x=y'(3xy^2-8y)\]
\[y'=\frac{x^3-2x}{3xy^2-8y}\] well that wasn't so bad was it
hope it is right
shall we check?

Answer 17

hmm, looks good!!

Answer 18

yeah too bad it is not the wolfram answer

Answer 19

ah. if only this wasn't an even numbered question ..

Answer 20

lets try again

Answer 21

thanks man!! I super appreciate your effort haha.

Answer 22

first of all i made a stupid algebra mistake

Answer 23

if we go ahead with \[(y^2)(xy-4)-x^2=0 \]then we could try to use product rule twice?

Answer 24

it will be a lot lot easier to multiply out first

Answer 25

\[xy^3-4y^2-x^2=0\] as a first step
then
\[y^3-3xy^2y'-8yy'-2x=0\]

Answer 26

then
\[y^3-2x=3xy^2y'+8yy'\]
\[y'=\frac{y^3-x}{3xy^2+8y}\] but that is still not right!

Answer 27

wolf has a \(-8y\) in the denominator where i have it as plus
but a totally different answer if you use hte quotient rule

Answer 28

hm, does wolf talk into account implicit differentiation?

Answer 29

i am going back to this
\[2yy'=\frac{(xy-4)\times 2x-x^2(y+xy')}{(xy-4)^2}\]

Answer 30

yes it does
it will find both \(\frac{dy}{dx}\) and also \(\frac{dx}{dy}\)

Answer 31

i did out question and got wolfram answer. 1 sec i will post picture

Answer 32

\[2yy'=\frac{(xy-4)\times 2x-x^2(y+xy')}{(xy-4)^2}\]
\[(xy-4)^2\times 2yy'=2x^2-8x-x^2y-x^3y'\]

Answer 33

http://www.wolframalpha.com/input/?i=y^2%3Dx^2%2F%28xy-4%29

Answer 34

OH MY.

Answer 35

\[2y(xy-4)^2y'+x^3y'=2x^2-8x-x^2y\]
\[y'=\frac{2x^2-8x-x^2y}{2y(xy-4)^2}\] is what i get

Answer 36

no another algebra mistake

Answer 37

\[y'=\frac{2x^2-8x-x^2y}{2y(xy-4)^2+x^3}\]

Answer 38

oh well, guess my algebra is not up to the task

Answer 39

it was a great effort!! I reward you.

Answer 40

actually i think it might be right

Answer 41

hopefully you can see that picture

Answer 42

yeah did you get that from wolf? i think they might be the same

Answer 43

wow, haha. thank you!!

Answer 44

i worked it out myself but it matches wolf answer

Answer 45

i wish i could give 2 rewards.. this question was too crazy

Answer 46

no worries. can you help me with my implicit differentiation question i have?

Answer 47

then you win
i can't find my mistake though...

Answer 48

guess i need pencil and paper too, stop trying to write it all here

Answer 49

@amyvincent12 what is your question?

Answer 50

y=csc(xy) find y' using implicit differentiation

Answer 51

\[y'=-\cos(xy)\cot(xy)(y+xy')\] is a start

Answer 52

i have that

Answer 53

ok i hope not, it is wrong
i meant
\[y'=-\csc(xy)\cot(xy)(y+xy')\]

Answer 54

then multiply out and get
\[y'=-y\csc(xy)\cos(xy)-xy'(\csc(xy)\cot(xy)\]

Answer 55

sorry, didn't see typo. yes, this is what i have.

Answer 56

put all the \(y'\) on the left and get
\[y'+xy'(\csc(xy)\cot(xy)=-y\csc(xy)\cot(xy)\]

Answer 57

factor out the \(y'\) to get
\[y'(1+x\csc(xy)\cot(xy))=-y\csc(xy)\cot(xy)\]

Answer 58

then divide and you are done

Answer 59

look reasonable?

Answer 60

i got an answer of \[y'=\frac{ -ycsc(xy)\cot(xy) }{1+xcsc(xy)\cot(xy)}\]

Answer 61

but doesn't match wolfram

Answer 62

me too

Answer 63

wolfram doesn't have denominator at all.

Answer 64

well, there are lots of trig identities you could use maybe to change it
let me check

Answer 65

the only trig identities that my prof wants us to change is sinx/cosx =tanx and cosx/sinx = cotx so thats probably my final answer just not for wolf

Answer 66

wolf gets the same answer i do

Answer 67

okay awesome maybe i made typo in wolf… thanks!!!!!

Answer 68

actually
when i type it in to wolf it gives
\[y'=y'(x) = -\frac{(y\cos(x y))}{\sin^2(x y)+x\cos(x y)}\] but when i got to "step by step" it gives my answer above
hmmm

Answer 69

probably some messing around with trig that get it, i don't see it

Answer 70

probably replace the 1 in the denominator by \(\sin^2(xy)+\cos^2(xy)\) and then some algebra