# How do i differentiate this?

1. Feb 19, 2004

### derivativeated

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

2. Feb 19, 2004

### 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

Last edited: Feb 19, 2004
3. Feb 20, 2004

### 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]

Last edited: Feb 21, 2004
4. Feb 21, 2004

### 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.

5. Feb 21, 2004

### matt grime

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

6. Feb 24, 2004

### HallsofIvy

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

7. Feb 24, 2004

### Zurtex

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.

8. Feb 24, 2004

### 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.