Can someone explain to me how to implicitly derive x^2-y^2= xy?
I got to 2x - 2y dy/dx = y+x.

Question

Can someone explain to me how to implicitly derive x^2-y^2= xy?
I got to 2x - 2y dy/dx = y+x.

parthkohli · Answer

why the y + x?

kaleidoscopicsink · Answer

Product rule gives you that when deriving the right side of the equation.

parthkohli · Answer

it doesn't say that the derivative of xy is x + y, does it?

kaleidoscopicsink · Answer

IT does though cause you do, CB + AD(I work in terms of A,B,C, and D) SO A is x, B is y C is the derivative of x so 1, and D is the derivative of y so 1. and when done in CB + AD you get y+x.

mathstudent55 · Answer

$\dfrac{d}{dx} uv = uv' + vu'$

kaleidoscopicsink · Answer

Yes, which is the same as my CB+AD.

mathstudent55 · Answer

The derivative of a product is the first times the derivative of the second plus the second times the derivative of the first.

mathstudent55 · Answer

The derivative of x with respect to x is 1.
The derivative of y with respect to x is y'.

mathstudent55 · Answer

$\dfrac{d}{dx} (xy) = xy' + yx' = xy' + y$

anonymous · Answer

Hi, maybe you don't know how to derive a phrase containing both x and y.

here's an example:
$$x ^{2}y \rightarrow \frac{ d(x ^{2}y) }{ dx }={ 2xy+x^2y' }$$

kaleidoscopicsink · Answer

I know how to derive with x and y. I just keep making such simple and stupid mistakes.

anonymous · Answer

Ok, then. go ahead then and put it here to be checked if you want.

mathstudent55 · Answer

Remember the power rule and the chain rule.

mathstudent55 · Answer

$x^2-y^2= xy$

$2x - 2yy' = xy' + yx'$

Do you understand it up to here?

kaleidoscopicsink · Answer

Yes, of course, I'm just struggling to do the y' part that you posted. I think the answer is (y-2x)/ (-2y-x).
And what mathstudent55 wrote about the product rule I have written done I just didn't post the y' part.
My equation looks like this in my notebook: 2x-2y(dy/dx)= y+x*(dy/dx)

mathstudent55 · Answer

Ok, here is the rest.

$xy' + 2yy' = 2x - y$

$y'(x + 2y) = 2x - y$

$y' = \dfrac{2x - y}{x + 2y} $

kaleidoscopicsink · Answer

2x − 2y
 2x/1+ 2y
y - 2x/ -2y - x
 0
Those are my options. I originally got what you are showing, but I thought it was wrong.

mathstudent55 · Answer

$y' = \dfrac{2x - y}{x + 2y}$

$y' = \dfrac{-1(2x - y)}{-1(x + 2y)}$

$y' = \dfrac{-2x + y}{-x - 2y}$

$y' = \dfrac{y-2x}{-2y -x }$

kaleidoscopicsink · Answer

Wow, so I unnecessarily confused myself, but had the right answer. Thank you everyone for all your help. cx And a huge thank you to @mathstudent55

mathstudent55 · Answer

You are very welcome.
You are correct. You did have the correct answer. You just had to make it look like one of the choices, so you could choose from the choices.