In the problem above, what is the difference between dy/dx and d/dx? Conceptually, how are they different. I know the former is the \Delta y with respect to \Delta x as x -> 0, but what is the latter?
And then my textbook does not use d/dx in this example. What gives?:
Here's another instance where they use the confusing d/dx notation:
On
I think what you are missing is that $\frac{dy}{dx}$ is the result of applying the operator $\frac{d}{dx}$ to $y$. $e^{sin(x)}cos(x)$ is the result of applying the operator $\frac{d}{dx}$ to $e^{sin(x)}$. If $y=e^{sin(x)}$, then, if you apply the same operator to both sides, you wind up with:
$$\frac{d}{dx}(y) = \frac{d}{dx}(e^{sin(x)})$$
This then simplifies to:
$$\frac{dy}{dx} = e^{sin(x)}cos(x)$$
Personally, I prefer separating $d()$ from $\frac{}{dx}$ as two separate operations, but nobody that I can find teaches it that way.
In the second example, $\theta$ is simply in the place of $x$, because $\theta$ is the variable the derivative is being taken with respect to. It's exactly the same operation.
See it this way: We are talking here about function terms in a real variable $x$, like $3x^2-7x+5$, $e^{\sin x}$, $\sqrt{1-x^2}$, etc.
The typographical picture ${d\over dx}$ denotes an operator that can be applied to such terms. It takes the derivative with respect to $x$ of such a term, according to the rules learnt in calculus 101.
Now in your context $y$ is an abbreviation for some more complicated term in the variable $x$. The author then does not write ${d\over dx}y$ in order to get the derivative, but he writes ${dy\over dx}$. That's all.
You have to be aware than in our "working analysis" we all are somewhat sloppy with the notations of variables, functions, operators, etc. The same $y$ can be an independent coordinate variable, a dependent variable tied to an independent variable $x$ via $y=f(x)$, or denote some given or unknown function taking values on the $y$-axis, and on and on.