If we have some closed form multivariable function say 2-in, 1-out, the cross sections of the graph parallel to the xz and yz plane have equations that are inherently closed form. But is the same true for the diagonal cross sections?

In a regression analysis, the variable that is being predicted is the "dependent variable."
a. Intervening variable 
b. Dependent variable 
c. None 
d. Independent variable

What is $R^{*}$ in math?

Repeated addition is called ?
A)Subtraction
B)Multiplication
C)Division

Multiplicative inverse of 1/7 is _?

Does the series converge or diverge this $\sum <!--\; \sum \; -->n!/n^n$

Use Lagrange multipliers to find the point on a surface that is closest to a plane.
Find the point on $z=1-<!--\; -\; -->2x^2-<!--\; -\; -->y^2$ closest to $2x+3y+z=12$ using Lagrange multipliers.
I recognize $z+2x^2+y^2=1$ as my constraint but am unable to recognize the distance squared I am trying to minimize in terms of 3 variables. May someone help please.

Just find the curve of intersection between $x^2+y^2+z^2=1$ and $x+y+z=0$

Which equation illustrates the identity property of multiplication? A$(a+\mathrm{bi})\times c=(\mathrm{ac}+\mathrm{bci})$ B$(a+\mathrm{bi})\times 0=0$ C$(a+\mathrm{bi})\times (c+\mathrm{di})=(c+\mathrm{di})\times (a+\mathrm{bi})$ D$(a+\mathrm{bi})\times 1=(a+\mathrm{bi})$

The significance of partial derivative notation
If some function like $f$ depends on just one variable like $x$, we denote its derivative with respect to the variable by:
$\frac{\mathrm{d}}{\mathrm{d}x}f(x)$
Now if the function happens to depend on $n$ variables we denote its derivative with respect to the $i$th variable by:
$\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}{x}_i}f(x_1,\cdots <!--\; \cdots \; -->,x_i,\cdots <!--\; \cdots \; -->,x_n)$
Now my question is what is the significance of this notation? I mean what will be wrong if we show "Partial derivative" of $f$ with respect to $x_i$ like this? :
$\frac{\mathrm{d}}{\mathrm{d}{x}_i}f(x_1,\cdots <!--\; \cdots \; -->,x_i,\cdots <!--\; \cdots \; -->,x_n)$
Does the symbol $\mathrm{\partial <!--\; \partial \; -->}$ have a significant meaning?

The function $f(x,y,z)$ is a differentiable function at $(0,0,0)$ such that $f_y(0,0,0)=f_x(0,0,0)=0$ and $f(t^2,2t^2,3t^2)=4t^2$ for every $t>0$. Define $u=(6/11,2/11,9/11)$, with the given about. Is it possible to calculate $f_u(1,2,3)$ or $f_u(0,0,0)$, or $f_z(0,0,0)$?

Given topological spaces $X_1,X_2,\dots <!--\; \dots \; -->,X_n,Y$, consider a multivariable function $f:\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}{X}_i\to <!--\; \to \; -->Y$ such that for any $(x_1,x_2,\dots <!--\; \dots \; -->,x_n)\in <!--\; \in \; -->\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}{X}_i$, the functions in the family $\{x\mapsto <!--\; \mapsto \; -->f(x_1,\dots <!--\; \dots \; -->,x_{i-<!--\; -\; -->1},x,x_{i+1},\dots <!--\; \dots \; -->,x_n){\}}_{i=1}^n$ are all continuous. Must $f$ itself be continuous?

Let $x$ be an independent variable. Does the differential dx depend on $x$?(from the definition of differential for variables & multivariable functions)

Let $f:M(n,\mathbb{R})\to <!--\; \to \; -->M(n,\mathbb{R})$ and let $f(A)=A{A}^t$. Then find derivative of $f$, denoted by $df$ .
So, Derivative of $f(df)$ if exists, will satisfy $limH\to <!--\; \to \; -->0\frac{||f(A+H)-<!--\; -\; -->f(A)-<!--\; -\; -->df(H)||}{||H||}=0$.

if $F(x,y)$ and $y=f(x)$,
$\frac{dy}{dx}=-<!--\; -\; -->\frac{\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}\left(F\right)}{\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}y}\left(F\right)}$
1) $F(x,y)$ 𝑎𝑛𝑑 $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$
2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constant

Let $f:{\mathbb{R}}^2\to <!--\; \to \; -->\mathbb{R}$ be defined as
$f(x,y)=\{\begin{array}{ll}(x^2+y^2)\cos <!--\; \; -->\frac{1}{\sqrt{x^2+y^2}},& \text{for }(x,y)\ne <!--\; \ne \; -->(0,0)\\ 0,& \text{for }(x,y)=(0,0)\end{array}$
then check whether its differentiable and also whether its partial derivatives ie $f_x,f_y$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for $f_x$ as function is symmetric in $y$ and $x$. So $f_x$ turns out to be
$f_x(x,y)=2x\cos <!--\; \; -->\left(\frac{1}{\sqrt{x^2+y^2}}\right)+\frac{x}{\sqrt{x^2+y^2}}\sin <!--\; \; -->\left(\frac{1}{\sqrt{x^2+y^2}}\right)$
which is definitely not $0$ as $(x,y)\to <!--\; \to \; -->(0,0)$. Same can be stated for $f_y$. But how to proceed with the first part?