Differentiation Variable

  • Thread starter womfalcs3
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  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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I'm really sure what you're trying to do, but if you have a function [tex]f(x,y)[/tex] and you want to know how [tex]f[/tex] changes in the limiting case with respect to a change in the ratio of [tex]y/x[/tex], then this is a case for the directional derivative. Note that [tex]v = (x, y)[/tex] is the vector where [tex]y/x[/tex] stays constant, so you'd want to take the vector orthogonal to that; ie, use [tex]v = (y, -x)[/tex]. In that case, [tex]\nabla_v f(x,y) = \nabla f(x,y) \cdot v[/tex]
 
  • #3
HallsofIvy
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Let v= u/y. Then, if f(u,y) is any function of u and y, [itex]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[/itex].

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

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

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