Exact Derivative Derivation I was given that:

The 'exact derivative' $df$ is the infinitesimal change in the value of function $f(x,y,z)$ when the independent variables change infinitesimally from $(x,y,z)$ to $(x+dx, y+dy, z+dz)$, given by
$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$

And asked to derive this result. My understanding of how this derivation may go is:

$f(x+dx,y+dy,z+dz)$
$\approx f(x,y,z) + df$
$=f(x,y,z) + \delta x +\delta y + \delta z\\
\approx f(x,y,z) + k_xdx + k_ydy+k_zdz\\$


So we have $f(x+dx,y,z) \approx f(x,y,z)+k_xdx\\
\implies k_x \approx \frac{f(x+dx,y,z)-f(x,y,z)}{dx}\\
\implies k_x = \lim_{dx \rightarrow 0} \frac{f(x+dx,y,z)-f(x,y,z)}{dx}\\
\implies k_x = \frac{\partial f}{\partial x}$


By similar argument obtain $k_y = \frac{\partial f}{\partial y}$ and $k_z = \frac{\partial f}{\partial z}$
Giving $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$

Is this OK? If not, what would be the correct way of deriving this result?
 A: When dealing with calculus on manifolds, given $f:M \to \Bbb R$, one has a very precise definition: $df_p:T_pM \to \Bbb R^n$ is given by $$df_p(v) = \frac{d}{dt}\bigg|_{t=0} f(\gamma(t)),$$where $\gamma:(-\epsilon,\epsilon)\to M$ satisfies $\gamma(0)=p$ and $\gamma'(0)=v$ --- one then proves that this doesn't depend on the choice of $\gamma$. Chosen coordinates $(x^1,\ldots,x^n)$ around $p$, one may write $$df_p = \sum_i a_i\,dx^i\big|_p,$$where $\{dx^1\big|_p,\ldots, dx^n\big|_p\}$ is a basis for the dual space $(T_pM)^*$. By linear algebra one has $$a_i = df_p\left(\frac{\partial}{\partial x^i}\bigg|_p\right),$$where $\{(\partial/\partial x^1)|_p,\ldots,(\partial/\partial x^n)|_p\}$ is the basis for $T_pM$ induced by the coordinate system. And from the way partial derivatives on a manifold are locally defined from a coordinate system, it just turns out that $$a_i = \frac{\partial f}{\partial x^i}(p).$$
Now, given that this exercise showed up in a calculus class that does not deal with precise geometric definitions of those objects, I would say your argument was very good. I do not see any room for improvement and would give you full marks for it.
