3 small error

A first answer to "how misleading": more than one will simplify and get $\frac{dy}{dx}=\frac{y}{x}$. A more serious objection is, thinking the derivative as the ratio of two infinitesimal increments $dy$ and $dx$ without the convenient foundation may lead a freshman student to the conclusion that every function is differentiable (if I can think to quantities $dy$ and $dx$, what's wrong in a harmless algebraic operation on them).

This does not mean one has to avoid $\frac{dy}{dx}$, but instead of using it to introduce the derivative "because it gives the right intuition", I would prefer a more rigorous definition, introducing the Leibnitz' notation only later, justifying it because it is formally consistent with the theorems about the derivatives of compositions and inverses of functions.

Personally, I prefer the definition via first order expansion: $f$ has derivative $m$ at $x$ if $f(x+h)=f(x)+mh+o(h^2)$ f(x+h)=f(x)+mh+o(h)$as$h\to 0$; as to the above mentioned composition rule, it is even more intuitive: the affine approximation of a composition is the composition of the affine approximations. (I happen to talk here on this point of view). 2 added 2 characters in body A first answer to "how misleading": more than one will simplify and get$\frac{dy}{dx}=\frac{y}{x}$. A more serious objection is, thinking the derivative as the ratio of two infinitesimal increments$dy$and$dx$without the convenient foundation may lead a freshman student to the conclusion that every function is differentiable (if I can think to quantities$dy$and$dx$, what's wrong in a harmless algebraic operation on them). This does not mean one has to avoid$\frac{dy}{dx}$, but instead of using it to introduce the derivative "because it gives the right intuition", I would prefer a more rigorous definition, introducing the Leibnitz' notation only later, justifying it because it is formally consistent with the theorems about the derivatives of compositions and inverses of functions. Personally, I prefer the definition via first order expansion:$f$has derivative$m$at$x$if$f(x+h)=f(x)+mh+o(h)$f(x+h)=f(x)+mh+o(h^2)$ as $h\to 0$; as to the above mentioned composition rule, it is even more intuitive: the affine approximation of a composition is the composition of the affine approximations. (I happen to talk here on this point of view).

A first answer to "how misleading": more than one will simplify and get $\frac{dy}{dx}=\frac{y}{x}$. A more serious objection is, thinking the derivative as the ratio of two infinitesimal increments $dy$ and $dx$ without the convenient foundation may lead a freshman student to the conclusion that every function is differentiable (if I can think to quantities $dy$ and $dx$, what's wrong in a harmless algebraic operation on them).
This does not mean one has to avoid $\frac{dy}{dx}$, but instead of using it to introduce the derivative "because it gives the right intuition", I would prefer a more rigorous definition, introducing the Leibnitz' notation only later, justifying it because it is formally consistent with the theorems about the derivatives of compositions and inverses of functions.
Personally, I prefer the definition via first order expansion: $f$ has derivative $m$ at $x$ if $f(x+h)=f(x)+mh+o(h)$ as $h\to 0$; as to the above mentioned composition rule, it is even more intuitive: the affine approximation of a composition is the composition of the affine approximations. (I happen to talk here on this point of view).