How to apply the chain rule to a double partial derivative of a multivariable function? f(x,y)=e^(xy) g(x,y)=f (sin(x^2+y),x^2+2y+1 Let’s compute d^2g/dx^2(0,0)

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?