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I know it's possible to do this:

$\frac{dy}{dx} \frac{dt}{dt} = \frac{dy}{dt} \frac{dt}{dx}$

but I wonder if this makes sense?

$\frac{d}{dx}\left(\frac{dt}{dt}\right) = \frac{d}{dt} \left(\frac{dt}{dx}\right)$

so if $t=x^4$ then $\displaystyle\frac{dt}{dx} = 4x^3$ and

$\frac{d}{dx}\left(\frac{dt}{dt}\right) = \frac{d}{dt} \left(4x^3\right)$

but this is like saying

$\frac{d}{dx} \left(1\right) = \frac{d}{dt} \left(4x^3\right)$

and $0=0$.

so it makes sense in that instance at least... I suppose at time this notation is still mysterious.

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    See also http://math.stackexchange.com/q/8961/1242.2011-06-22

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The notation itself is just that, a notation. While it may seem 'clean and safe' to move differentials around as though they're just another variable, technically speaking it isn't a valid way to do mathematics.

Now, provided you understand and follow the rules, some manipulations along those lines can still arrive at a correct result albeit through potentially dubious means.

A good example of this shows up commonly in differential equations texts:

$f(x,y)dx + g(x,y)dy = 0$

... which, when written in proper form is:

$\frac{dy}{dx} = -\frac{f(x,y)}{g(x,y)}$

However, when used as a mnemonic device the former is a good way to help a student remember how to find the adjoint of the ODE and ultimately arrive at a general solution.

Under the hood, the former can be re-written as:

$f(x,y)\frac{dx}{dt} + g(x,y)\frac{dy}{dt} = 0$

... due to the chain rule:

$\frac{dy}{dt}\frac{dt}{dx} = -\frac{f(x,y)}{g(x,y)}$

... where here it must be assumed that $x(t)$ has an inverse $x^{-1}(t) = t(x)$ such that $\displaystyle\frac{1}{\frac{dt}{dx}} = \frac{dx}{dt}$ in order to arrive at the homogeneous equation. I'm probably missing some other pertinent details, but this is closer to a more correct way to work with differentials than the ad-hoc methods usually taught in differential equations texts.

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    @Noteventhetutorknows - The fractional notation helps give the notion of rates, but it also gives the impression you can just do normal algebraic manipulations on them -- which is definitely not true. It can be done, as Americo Tavares points out, but it must be done very carefully and rigorously.2011-06-22
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The differential of a differentiable function $f(x)$ at $x_0$ is the expression f'(x_0)dx. If $f(x)=x$, then f'(x_0)dx=1\cdot dx=dx. The algebraic manipulation of differentials may be intuitive in certain cases but it has to be checked, i. e. proved or disproved rigorously.