How Can I Calculate dy/dx for x^2 - xy - y^2 = 3?

[derivativeated]

x^2 - xy - y^2 = 3
how do i work out dy/dx?

 

[matt grime]

either make y a function of x, which is trival but dull or just differentiate everything wrt x and use the chain rule and product rule.

example, one you can do anyway, suppose xy=1

then assuming this implicitly defines a function y in terms of x (notice that word implicit, look in your notes for a reference to it)

then \frac{d}{dx}xy = \frac{d}{dx}1 that is to say (\frac{d}{dx}x)y + x(\frac{d}{dx}y) = 0. of course dx/dx=1 so, y+ x\frac{dy}{dx}=0 and we see that \frac{dy}{dx} = -y/x recall that y= 1/x, and we see \frac{dy}{dx} = -1/x^2 try and apply this idea to you example

 

[Orion1]

Implicit Differentiation...


differentiate:
x^2 - xy - y^2 = 3

Implicit Differentiation:
(d/dx)[x^2 - xy - y^2] = (d/dx)[3]

(d/dx)[x^2] - (d/dx)[xy] - (d/dx)[y^2] = (d/dx)[3]

[/color]

 

[Zurtex]

I'm a little confused with the answers above. When differentiating some function of y with respect to x, is it not simply the derivate of the function with respect to y multiplies by the derivative of y with respect to x?

Such that:

x^2 - xy - y^2 = 3

2x - \left( y + x \frac{dy}{dx} \right) - 2y \frac{dy}{dx} = 0

2x - y - x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0

x \frac{dy}{dx} + 2y \frac{dy}{dx} + y - 2x = 0

(x + 2y) \frac{dy}{dx} = 2x - y

\frac{dy}{dx} = \frac{2x - y}{x + 2y}

Please say if this is wrong somehow, I need the practise.

 

[matt grime]

that seems to be a correct assertion. this is called implicit differentiation.

 

[HallsofIvy]

Originally posted by Zurtex
I'm a little confused with the answers above. When differentiating some function of y with respect to x, is it not simply the derivate of the function with respect to y multiplies by the derivative of y with respect to x?

Such that:

x^2 - xy - y^2 = 3

2x - \left( y + x \frac{dy}{dx} \right) - 2y \frac{dy}{dx} = 0

2x - y - x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0

x \frac{dy}{dx} + 2y \frac{dy}{dx} + y - 2x = 0

(x + 2y) \frac{dy}{dx} = 2x - y

\frac{dy}{dx} = \frac{2x - y}{x + 2y}

Please say if this is wrong somehow, I need the practise.

Yes, that was, in fact, exactly what matt grime orginally said!

 

[Zurtex]

Originally posted by HallsofIvy
Yes, that was, in fact, exactly what matt grime orginally said!

I didn't really understand the method though and got a bit confused so I wanted to check I knew how to do it, I was unsure why the example of if xy = 1 was given.

Probably just the way I look at maths at the level I do it, I presume it would differ when I go to university.

 

[matt grime]

I gave that example for two reasons: it was easy to rearrange and solve without implicit differentiation, so that you could see that you got the answer you thought you ought to get, and because I didn't want to just solve your homework problem for you, but to prompt you into trying it again for yourself, changind the details as necessary.