Question

use implicit differentiation to find dy/dx.
2xy-y^2=1. answers are: x/(y-x), x/(x-y), y/(y-x), or y/(x-y).

Answer 1

Product rule for 2xy
Chain rule for y^2

know those rules already ?

Answer 2

yes, but could you show your steps, please? They're new to me...

Answer 3

let me give you an example instead,

say if i have \(x^2*e^y\)
using product rule, its derivative will be
\((x^2)' e^y + x^2 (e^y)' \\ = 2x e^y + x^2 e^y \dfrac{dy}{dx}\)

see if you get this.
i have also used chain rule for 2nd term.

Answer 4

if you get that, then derivative of xy will be as easy as eating an icecream :P :)

Answer 5

so i would
2x(dy/dx)+y(2)-2y=0?

Answer 6

"2x(dy/dx)+y(2)" is exactly correct for the derivative of 2xy!

for the derivative of y^2, we will need chain rule.
like derivative of y^8 = 8y^7 dy/dx

Answer 7

oh, i left off the dy/dx! so the derivative for y^2 would be 2y(dy/dx)?

Answer 8

yes! :)
2x(dy/dx)+y(2)-2y(dy/dx)=0
now just isolate dy/dx

Answer 9

so subtract 2y to the left and factor dy/dx.
(dy/dx)(2x-2y)=-2y
then divide both sides by 2x-2y
which equals -y/x-y. then i multiply by -1/-1 and i get
y/y-x. right?

Answer 10

it surely is correct :)
good work!

Answer 11

Thank you so much!

Answer 12

most welcome ^_^