This is a continuation of Some basic doubts about partial differentiation.
Doubt #3 : : Let us say that we have $f=f(x_1,x_2,....,x_n)$ where $x_i \in R$. Is the following equation always true ?
$$df=\frac {\partial f}{\partial x_1}dx_1+....+\frac {\partial f}{\partial x_n}dx_n$$
Doubt #4 : : Let $h=h(x,y,z)$ where $z$ itself is a function of the independent variables $x$ and $y$, i.e., $z=z(x,y)$.
Which of the following equations is true ?
$$ dh=\frac{\partial h}{\partial x}dx + \frac{\partial h}{\partial y}dy + \frac{\partial h}{\partial z}dz\,\,\,\,(1)$$
$$dh=\frac{\partial h}{\partial x}dx + \frac{\partial h}{\partial y}dy \,\,\,\,(2)$$
I wrote Equation $(1)$ thinking that $h$ depends on $x, y$ and $z$. I wrote Equation $(2)$ thinking that because $z$ can be expressed in terms of $x$ and $y$, $h(x,y,z)$ is basically a function of only $x$ and $y$. But I can not decide which of these two equations is actually correct.
$\frac {\partial h}{\partial z}$ means differentiating $h$ wrt $z$ while holding $x$ and $y$ constant. But if $x$ and $y$ are constant, doesn't that mean that $z=z(x,y)$ is also a constant ? Does $\frac {\partial h}{\partial z}$ even mean anything here ?
NOTE : As you can clearly see by the nature of my doubts, I am just a beginner in multivariable calculus. So please answer in simple terms. Thanks in advance :-).
With GReyes's help, I think I have found an answer to Doubt #4 :
We know that $h=h(x,y,z)$ and $z=z(x,y)$. Equation $(1)$ (see question) is valid if we have not eliminated the variable $z$ by using $z=z(x,y)$. What I mean by "elimination" is that if the dependence of $z$ on $x$ and $y$ is already given , i.e , if the function $z(x,y)$ is already given, then we can simply put substitute $z=z(x,y)$ in the expression for $h(x,y,z)$ and thus the function now becomes $h(x,y,z(x,y)) = g(x,y)$ (say). The new function $g(x,y)$ is just $h$ but with $z$ eliminated. So if $z$ is not eliminated, then equation (1) is correct. If $z$ has been eliminated, then equation (2) is correct but a small correction must be taken into account :
$$dh=\frac{\partial g}{\partial x} dx + \frac {\partial g}{\partial y} dy \,\,\,\,(2*)$$
If you compare $(2)$ with $(2*)$, then it is clear that $h$ must be replaced with $g$ because $h$ would still have the variable $z$ in its expression whereas $g$ won't.
To summarize :
$$dh=\frac{\partial h(x,y,z)}{\partial x} dx+\frac{\partial h(x,y,z)}{\partial y} dy+\frac{\partial h(x,y,z)}{\partial z} dz=\frac{\partial g(x,y)}{\partial x} dx + \frac {\partial g(x,y)}{\partial y} dy$$