Proper notation to denote "take the derivative of this" I was working on a related rates calculus problem on my math homework. The point of the problem is to start with the area formula of a triangle, $A= \frac 12 b\dot{}h$ and then to take the derivative. Since base, height, and area are all changing variables in this word problem, I treated A, b and h as variables, not constants and took the derivative of all three and used the product rule for the right side of the equation.
$$dA = \frac 12 (b \ dh + h \ db)$$
I know how to take the derivative, but not how to mathematically write "take the derivative of this function". I know that $y'$ and $\frac{dy}{dx}$ is the derivative of the function and if I had the equation $y = x^2$
then I would write $[\frac{d}{dx}] y =[ \frac{d}{dx} ]x^2$ to say "take the derivative of the function". But what if I have additional variables, how would you write "take the derivative" in math terms? And how would all of the dx's dy's, etc. cancel out?
 A: It sounds to me like you're saying that $A,b,h$ are all changing with respect to some other variable (let's call it $t$ for time), in which case I'd say you took the derivative with respect to $t$ of both sides, writing $\frac{dA}{dt} = \frac{d}{dt}\left[\frac{1}{2}bh\right] = \frac{1}{2}\left(b\frac{dh}{dt} + h\frac{db}{dt}\right)$. The equation you wrote seems to re-express this using infinitesimals to "cancel the $dt$'s".
A: The more useful concept here is the differential. $\mathrm{d}A$ is the differential of $A$ (it is not unreasonable to call it the "derivative" of $A$)
Differentials satisfy all of the algebraic properties you expect derivatives to have. Also, if $f$ is a differentiable univariate function, then $\mathrm{d}f(z) = f'(z) \mathrm{d}z$, no matter what $z$ is.
In your case, you have correctly computed the differential of $A$: you do indeed have
$$ \mathrm{d}A = \frac 12 (b \ \mathrm{d}h + h \ \mathrm{d}b) $$
Depending on the problem, one might continue on to do things like:


*

*Set one of the differentials to zero, because that variable is being held constant

*Do additional computations on $\mathrm{d}h$ and $\mathrm{d}b$. For example, you may have $h$ written as a function of $b$, or both as functions of some third variable $t$

*Solve for one of the other differentials

*If you arrive at a formula that expresses one differential as a multiple of another differential, then taking the ratio gives the usual derivative (i.e. as you learned in introductory calculus) that the notation suggests.

