Math question

asdf1
why does (-y/x^2)dx+(1/x)dy=d(y/x)?

Homework Helper
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
$$\frac{\partial}{\partial y}(\frac{y}{x})=\frac{1}{x}$$

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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?

asdf1
it's hard to understand...

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

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.

Homework Helper
asdf1 said:
it's hard to understand...

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.

Homework Helper
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.

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asdf1
thank you! :) that clears a lot of things up~