Question

find where the slope of the curve is defined:
x^2y-xy^2=4

Answer 1

does this mean "find the derivative, then check to see where the denominator is zero"?

Answer 2

start with
\[2xy+x^2y'-y^2-2xyy'=0\] solve for \(y'\) you will get a fraction
then see for which \(x\) and \(y\) the denominator will be zero

Answer 3

if that is not clear or if you need more help let me know

Answer 4

sorrry I dont get it. Can you please explain the steps please?

Answer 5

sure
this is an implicit differentiation problem
do you know how to find \(y'\) if \(x^2y-xy^2=4\) ?

Answer 6

No I dont..

Answer 7

ok then that is where we need to start
you are thinking in this equation, that represents some curve, that \(y\) is a function of \(x\) even though we cannot solve for \(y\)
in other words we are just imagining \(y\) as a function of \(x\) i.e \(y=f(x)\)

Answer 8

now we can think of x^2y-xy^2=4
\[x^2y-xy^2=4\] instead as
\[x^2f(x)-xf^2(x)=4\] and take the derivative of both sides
the derivative of \(4\) is 0, so we will get a 0 on the right.
the derivative on the left requires both the product rule and the chain rule
so when we take the derivative of \(f^2(x)\) we will get \(2f(x)f'(x)\) by the chain rule

Answer 9

using the product rule we get that the derivative is
\[2xf(x)+x^2f'(x)-f^2(x)-2xf(x)f'(x)=0\] of course this is just an explanation, it is a lot easier to write
\[2xy+x^2y'-y^2-2xy'=0\]
let me know which if any step was confusing

Answer 10

Okay I understand this part now.. but what do I do next?

Answer 11

now that we have \[2xy+x^2y'-y^2-2xy'=0\] we need to solve for \(y'\)

Answer 12

not too much here just some algebra
\[x^2y'-2xy'=y^2-2xy\]
\[(x^2-2x)y'=y^2-2xy\]
\[y'=\frac{y^2-2xy}{x^2-2x}\]
check my algebra

Answer 13

and this will be undefined if the denominator is zero, so you last job, assuming that my answer is correct, is to solve
\[x^2-2x=0\] which you can probably do with your eyeballs

Answer 14

Thank you so much! I get it now. :)

Answer 15

oh crap my algebra is wrong!!!

Answer 16

hold on lets get it right, i made an algebra mistake

Answer 17

alright

Answer 18

we start with
\[2xy+x^2y'-y^2-2xyy'=0\] after taking the derivative using implicit differentiation, in this case in the form of the product and chain rule
now we solve for \(y'\) carefully
\[2xy+x^2y'-y^2-2xyy'=0\]
\[x^2y'-2xyy'=y^2-2xy\]
\[(x^2-2xy)y'=y^2-2xy\]
\[y'=\frac{y^2-2xy}{x^2-2xy}\] that looks better

Answer 19

now to solve
\[x^2-2xy=0\]
\[x(x-2y)=0\] so \(x=0\) or \(x=2y\) will make this undefined

Answer 20

Thanks again!

Answer 21

you can ignore \(x=0\) because it is not in the domain of this expression

Answer 22

if you replace \(x\) by \(0\) you get \(0=4\) in the original equation, so \(x=0\) is not on the curve anyway
it is the other part that is your answer

Answer 23

and of course you are still not done, because now that you have \(x=2y\) you have to solve in the original equation to see exactly what \(x\) and \(y\) are

Answer 24

that is you have to solve
\[x^2y-xy^2=4\] if \(x=2y\) you get
\[4y^3-2y^3=4\]
\[2y^3=4\]
\[y^3=2\]
\[y=\sqrt[3]{2}\]

Answer 25

and you are STILL not done because you also have to solve for \(x\) but i will let you do that on your own.
should be easy since you already have \(y\)

Answer 26

Okay . Thank you so much! :)