multi-variable notation for continuous functions. $f(x,y) = \cos(x^2y) $
we know that cosine is continuous on $ \Bbb R $ that $x$ is continuous on $ \Bbb R $ and that $y$ is continuous on $ \Bbb R $ we also have a theorem that says composition of functions is continuous and products of functions is continuous. i want to write this out in a way that makes sense. but im not sure what makes sense.
let $h(x)= x$ let $g(y)=y $ let $p(x)= \cos(x) $ 
$h(x),g(y), p(x)$ are continuous functions
Then $h(x)*h(x) $ is continuous on $\Bbb R$ and $(h(x)*h(x))* g(y) $ is continuous on $\Bbb R $ and $p[(h(x)*h(x))*g(y)] $ is continuous on $ \Bbb R $ and hence 
$f(x,y) = p[(h(x)*h(x))*g(y)]$ is continuous on? 
do i want to somehow write something like $h(x,y)=x$ instead and make the same argument? 
 A: Actually you have to deal with $h(x,y)=x^{2}$ and $g(x,y)=y$ and then $p(u)=\cos u$ because of that $f(x,y)=p\circ(h\cdot g)(x,y)$.
Note that if we use only that $h(x)=x^{2}$, the domain of $h$ is only "one-dimensional", which has nothing to do with the "two-dimensional" domain of $f$.
A: Personally, I would note that
$$ m : \mathbb{R}^2 \to \mathbb{R} 
\qquad\text{defined by}\qquad
m(a,b) = ab $$
is continuous (i.e. multiplication is a continuous).  From this, we can conclude that
$$ s : \mathbb{R}^2 \to \mathbb{R}^2
\qquad\text{defined by}\qquad
s(a,b) = (a^2,b)
$$
is also continuous.  But then
$$ \cos(x^2 y)
= \cos( m(s(x,y)) )
= (\cos \circ m \circ s) (x,y). $$
This is then a composition of continuous functions, and is therefore continuous.  The domain of this function is $\mathbb{R}^2$, as it requires two real inputs (and, by the way, has one real output).
A: Yes—because your final function takes two arguments $x$ and $y$, it helps to build it out of parts that all take two arguments. For this purpose,


*

*Consider the projections $p(x,y) = x$ and $q(x,y)=y$. We can easily show that these functions are continuous.

*Hence the product $h(x,y) \equiv p(x,y) \cdot p(x,y) \cdot q(x,y) = x^2y$ is continuous.

*Finally, the function $\cos$ is continuous. Hence the function
$$f(x,y) = \cos(h(x,y)) = \cos{(x^2y)}$$
is continuous.

