# Math question

1. Aug 5, 2005

### asdf1

why does (-y/x^2)dx+(1/x)dy=d(y/x)?

2. Aug 5, 2005

### lurflurf

because
$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$
and
$$\frac{\partial}{\partial x}(\frac{y}{x})=\frac{-y}{x^2}$$
and
$$\frac{\partial}{\partial y}(\frac{y}{x})=\frac{1}{x}$$

Last edited: Aug 5, 2005
3. Aug 5, 2005

### asdf1

but how do you think of going from the left side to the right side if you haven't seen the right side before?

4. Aug 6, 2005

### asdf1

it's hard to understand...

5. Aug 7, 2005

### Staff: Mentor

$$\frac{d}{dx}(\frac{y}{x})= \frac{d}{dx} (y*\frac{1}{x})$$

If one has done the derivative of a product, then one knows -

$$\frac{d}{dx}(u(x)*v(x))= v(x) * \frac{d}{dx} u(x) + u(x) * \frac{d}{dx} v(x)$$

This can be verified with the definition of the derivative.

Then $$\frac{d}{dx}(\frac{1}{x})= -\frac{1}{x^2}$$,

which can also be shown using the definition of the derivative.

6. Aug 7, 2005

### matt grime

no it isn't, it's a definition. if you don't know what the definitions are then of course it will seem hard, but this is exactly what the notation means.

7. Aug 7, 2005

### HallsofIvy

Staff Emeritus
Yes, it is clear that $$d\frac{y}{x}= -\frac{y}{x^2}dx+ \frac{1}{x}dx$$ because $$\frac{d(\frac{a}{x})}{dx}= -\frac{a}{x^2}$$ for any constant a and $$\frac{d(\frac{y}{a})}{dy}= \frac{1}{a}$$ for any constant a.

But how would you go from $$-\frac{y}{x^2}dx+ \frac{1}{x}dx$$ to $$d{(\frac{y}{x})}$$? That, I think, is the question being asked and I'll bet it's in your textbook: look under "exact differentials" or "integrals independent of path". (Actually, going that direction gives a more general answer- you have a "constant of integration".)

You know that $$dF(x,y)= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y} dy$$. So whatever F is, if $$dF= -\frac{y}{x^2}dx+ \frac{1}{x}dx$$ then we must have $$\frac{\partial F}{\partial x}= -\frac{y}{x^2}$$ and $$\frac{\partial F}{\partial y}= \frac{1}{x}$$.

In particular, taking the anti-derivative of $$\frac{\partial F}{\partial x}= -\frac{y}{x^2}$$ , we have (remembering that the partial derivative treats y like a constant), $$F(x,y)= \frac{y}{x}+ g(y)$$ because the anti-derivative of $$\frac{a}{x^2}= ax^{-2}$$ is -ax-1 and here a is -y. Notice that g(y)! That's the "constant" of integration- no matter what function of y g is, taking the partial derivative with respect to x is 0.

Now, differentiate that with respect to y: $$\frac{\partial \frac{y}{x}+ g(y)}{\partial y}= \frac{1}{x}+ g'(y)$$ (Since g is a function of y only, the derivative of g is an ordinary derivative). Now compare that with $$\frac{\partial F}{\partial y}= \frac{1}{x}$$. We must have $$\frac{1}{x}+ g'(y)= \frac{1}{x}$$ which just tells us that g'(y)= 0 so g(y) really is a constant!

$$F(x,y)= \frac{y}{x}+ C$$ or $$d(\frac{y}{x}+C)= -\frac{y}{x^2}dx+ \frac{1}{x}dx$$. Of course, your example is taking C= 0.

Last edited: Aug 7, 2005
8. Aug 8, 2005

### asdf1

thank you! :) that clears a lot of things up~