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What's most misleading about Leibnizian notation is its implicit context dependence. After you get over that hurdle, it will be easy to safely think of $dy/dx$ as a fraction.

In the context of $y=f(x)$, you think of $dx$ either as an arbitrary nonzero infinitesimal also called $\Delta x$---I did this, using Keisler's book last fall---or as a nonzero real $\Delta x$ small enough for whatever your accuracy you currently need. Either way, $dy$ is defined as $f'(x)dx$, where $f'(x)$ is defined as the usual limit of difference quotients $\Delta y/\Delta x$. Of course, in the $x=g(y)$ context, the meanings of $dx$ and $dy$ switch, as do the meanings of $\Delta x$ and $\Delta y$. In the $z=h(x,y)$ context, the meanings of $dx$, $dy$, $\Delta x$, and $\Delta y$ change yet again.

The "small enough, but not infinitely small" approach is what you'll find in standard calculus textbooks, with a section devoted to the distinction between $\Delta y$ and $dy$ (in the $y=f(x)$ context).

That said, this fall I'm planning to de-emphasize $dy/dx$ as much as I can get away with. Whether I use the little-o notation or not, I will push hard (with lots of numerical examples) on the $\Delta y=f'(x)\Delta x+o(\Delta x)$ definition of $f'(x)$, and how this makes the chain rule true but not trivial.

If $y=x^2$ y=f(x)=x^2$and$dx=\Delta x$is small (but not infinitely small this time around), then$\Delta(x^2)$equals$(x+\Delta x)^2-x^2$equals$2x\Delta x+\Delta x^2$equals$2x\Delta x+(\mathrm{small})\Delta x$, so$dy=2x\ dx$and$f'(x)=2x$. In the context of$y=f(u)$and$u=g(x), my presentation of the chain rule will just be that a first-order approximation of a first-order approximation is a first-order approximation: \begin{align*} \Delta y&=f'(u)\Delta u+(\mathrm{small}_1)\Delta u\\ &=f'(u)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)+(\mathrm{small}_1)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)\\ &=f'(u)g'(x)\Delta x+(\mathrm{small})\Delta x \end{align*} No fractions here! 1 [made Community Wiki] What's most misleading about Leibnizian notation is its implicit context dependence. After you get over that hurdle, it will be easy to safely think ofdy/dx$as a fraction. In the context of$y=f(x)$, you think of$dx$either as an arbitrary nonzero infinitesimal also called$\Delta x$---I did this, using Keisler's book last fall---or as a nonzero real$\Delta x$small enough for whatever your accuracy you currently need. Either way,$dy$is defined as$f'(x)dx$, where$f'(x)$is defined as the usual limit of difference quotients$\Delta y/\Delta x$. Of course, in the$x=g(y)$context, the meanings of$dx$and$dy$switch, as do the meanings of$\Delta x$and$\Delta y$. In the$z=h(x,y)$context, the meanings of$dx$,$dy$,$\Delta x$, and$\Delta y$change yet again. The "small enough, but not infinitely small" approach is what you'll find in standard calculus textbooks, with a section devoted to the distinction between$\Delta y$and$dy$(in the$y=f(x)$context). That said, this fall I'm planning to de-emphasize$dy/dx$as much as I can get away with. Whether I use the little-o notation or not, I will push hard (with lots of numerical examples) on the$\Delta y=f'(x)\Delta x+o(\Delta x)$definition of$f'(x)$, and how this makes the chain rule true but not trivial. If$y=x^2$and$dx=\Delta x$is small (but not infinitely small this time around), then$\Delta(x^2)$equals$(x+\Delta x)^2-x^2$equals$2x\Delta x+\Delta x^2$equals$2x\Delta x+(\mathrm{small})\Delta x$, so$dy=2x\ dx$and$f'(x)=2x$. In the context of$y=f(u)$and$u=g(x)\$, my presentation of the chain rule will just be that a first-order approximation of a first-order approximation is a first-order approximation:

\begin{align*} \Delta y&=f'(u)\Delta u+(\mathrm{small}_1)\Delta u\\ &=f'(u)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)+(\mathrm{small}_1)(g'(x)\Delta x+(\mathrm{small}_2)\Delta x)\\ &=f'(u)g'(x)\Delta x+(\mathrm{small})\Delta x \end{align*}