General clarification for derivative notation I am a bit confused on the different notations of derivatives, could you help me clear it up?
The following can be interpreted as:


*

*the total derivative of f wrt x, or equivalently, the derivative of f(x) wrt x

*the partial derivative of f wrt x, or equivalently, the derivative of f(x,y,z) wrt x


$\dfrac{df}{dx}=\dfrac{d(f(x))}{dx},\, f=f(x)$
$\dfrac{\partial f}{\partial x}=\dfrac{\partial(f(x,y,z))}{\partial x},\, f=f(x,y,z)$
now the above is different from the below, which is:


*

*the total derivative of f wrt x, evaluated at the point a

*the partial derivative of f wrt x, evaluated at the point a, b, c


$\dfrac{df}{dx}(a)=\dfrac{d(f(x))}{dx}(a),\, f=f(x)$
$\dfrac{\partial f}{\partial x}(a,b,c)=\dfrac{\partial(f(x,y,z))}{\partial x}(a,b,c),\, f=f(x,y,z)$
however if we do a super contrived example and set a = x and a, b, c=x, y, z then the following equality holds (except super contrived, yes?)
$\dfrac{df}{dx}=\dfrac{d(f(x))}{dx}=\dfrac{d(f(x))}{dx}(x),\, f=f(x)$
$\dfrac{\partial f}{\partial x}=\dfrac{\partial(f(x,y,z))}{\partial x}=\dfrac{\partial(f(x,y,z))}{\partial x}(x,y,z),\, f=f(x,y,z)$
no, this is not how I normally write derivatives, and I am only bending the rules so far as to test the boundaries/semantics

Is the preceding interpretation correct?
 A: The idea behind what you wrote should be correct, but 


*

*The notation you use is not the most standard.

*In weird edge cases, what you wrote won't match up perfectly with what a calculus textbook or similar may write.


For point 1., instead of $\dfrac{\partial\left(f\left(x,y,z\right)\right)}{\partial x}\left(a,b,c\right)$, I think it would be more common, and almost certainly clearer, to write something like $\left.\dfrac{\partial\left(f(x,y,z)\right)}{\partial x}\right|_{(x,y,z)=(a,b,c)}$.
For point 2., if everyone is being a careful explicit mathematician, or if we're in a common case, then I think the answer to your question is "yes", the partial derivative is just "the derivative at an arbitrary point $(x,y,z)$". However, this may have an exception in a weird case. MathInsight has the sort of example I have in mind, although I've seen it elsewhere as well.
Consider the function $f(x,y)=\dfrac{x^3+x^4-y^3}{x^2+y^2}$ extended to be continuous on the whole plane. Equivalently, $f\left(x,y\right)=\begin{cases}\dfrac{x^{3}+x^{4}-y^{3}}{x^{2}+y^{2}} & \text{ where defined}\\0 & \text{ if }\left(x,y\right)=\left(0,0\right)\end{cases}$. Then someone might write use some differentiation rules and write something like $\dfrac{\partial f}{\partial x}=\dfrac{x^4+2x^5+3x^2y^2+4x^3y^2+2xy^3}{\left(x^2+y^2\right)^2}$. But this is undefined at $(0,0)$, even though the $x$-partial at $(0,0)$ is defined and equals $1$ (in symbols: $\left.\dfrac{\partial f}{\partial x}\right|_{(x,y)=(0,0)}=1$). 
