# Differentiation Variable

1. May 2, 2009

### womfalcs3

I'm working out some problems, and I'm ending up with a term similar to the following:

du/d(y/u)

I'm differentiating with respect to y/u. Both y and u are variables. How can I divide that up to represent differentiation with just one variable (Even if it means expanding the term)?

Is it mathematically possible to do that?

Thanks.

2. May 2, 2009

### junglebeast

I'm really sure what you're trying to do, but if you have a function $$f(x,y)$$ and you want to know how $$f$$ changes in the limiting case with respect to a change in the ratio of $$y/x$$, then this is a case for the directional derivative. Note that $$v = (x, y)$$ is the vector where $$y/x$$ stays constant, so you'd want to take the vector orthogonal to that; ie, use $$v = (y, -x)$$. In that case, $$\nabla_v f(x,y) = \nabla f(x,y) \cdot v$$

3. May 3, 2009

### HallsofIvy

Staff Emeritus
Let v= u/y. Then, if f(u,y) is any function of u and y, $df/dv= \partial f/\partial u \partial u\partial v+ \partial f/\partial y \partial y/partial v[itex]. Since, here, v= u/y, so u= yv and [itex]\partial u/\partial v= y[itex]. Similarly, y= u/v so [itex]\partial y/\partial v= -u/v^2= -u/(u^2/y^2)= -y^2/u$.

That is, $df/dv= y\partial f/\partial u- (y^2/u)\partial f/\partial y$

And, since here f(u,y)= u, $\partial f/\partial u= 1$ and $\partial f/\partial y= 0$ so we have df/dv= y.