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5

To clear up confusion from this, when I teach I make a massive emphasis of the fact that a function takes in an ordered list of inputs (and not something like 'variables'). Then I use notation like $D_1f$ to denote the partial derivative of $f$ with respect to its first input, $D_2f$ the second input etc. I go on about how each of these is a new function ...


2

One can fix this problem in another way - by expressing partial derivatives in a more natural form not involving quotients*. Given $$f = xyt$$ we have $$df = yt\,dx + xt\,dy + xy\,dt$$ no matter how the variables $x$, $y$, and $t$ may or may not be related - and this simply reads: The rate of change in $f$ is $yt$ times the rate of change in $x$ plus $xt$ ...


5

There is, as others have said, nothing wrong with $$\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t} \frac{dt}{dt}$$ Note that a partial derivative like $\frac{\partial f}{\partial t}$ is really to be understood as the derivative of $f$ with respect to its third variable,...


11

I guess you are aware of it, but since you don't say so explicitly: the chain rule on the top of your post does not apply to your example, since that $f$ is not a function of $x,y$ but of $x,y,t$. Also, as written, you are confounding the modern concept of "function" $f:\mathbb{R}^3\to\mathbb{R}$ with the original notion of "function of". ...


10

Do you ever introduce the Jacobian matrix, or the derivative of a function at a point as a linear map? This clarifies everything. If not, and you must restrict yourself to "traditional multivariable calculus" notation, you could write the chain rule as $(f \circ \gamma)'(t) = \nabla f|_{\gamma(t)} \cdot \gamma'(t)$ In your example $f(x,y,t) = xyt$ ...


16

What's wrong with this?: $$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t} \frac{dt}{dt}$$ with $\frac{dt}{dt}=1$. It's what you would get if you had $f(x,y,z)$, except that $z=t$. You do have to be a bit careful, though. You want to be clear that $f(x,t,t)=xt^2$ ...


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