if $u=x^2$ can you say $x^2$ is a function of $u$ If $$u=x^2$$ then obviously $u$ is a function of $x$ and $x$ is not a function of $u$.
But can we say that $x^2$ is a function of $u$?
Thanks
 A: We can not say that $x$ is a function of $u$. By definition of a function, a function $f$ from a set $X$ to a set $Y$ is defined by a set $G$ of ordered pairs $(x,y)$ with $x\in X$ and $y\in Y$ such that for every $x\in X$, there is only one $y\in Y$. Now, we can see that $x$ is not a function of $u$. We can see that for example $u=4$ has two different $x$ values, namely $x=2$ and $x=-2$. This definition is equivalent with saying $u:\mathbb{R}\to\mathbb{R};x\to x^2$ is not a bijection. If we define $u:[0,\infty)\to[0,\infty);x\to x^2$, we can inverse $u$ such that we can say that $x$ is a function of $u$.
A: The answer is yes. It is a simple exercise in substitution. Do $t = x^2$ and you see that:
$$t = u$$
And you see $t$ is a function of $u$, actually it is probably the simplest function one can think of. It is like $f(x) = x$ in first precalc course.

One example where it is common to mention functions of other functions like this is the trigonometric polynomials which due to the trigonometric laws have special properties. For example:
$$f(x)=\sin(x)^2 - 2\sin(x)+1$$
Which we can see that after doing the substitution $$t=\sin(x)$$
It becomes a polynomial:
$$t^2 - 2t + 1$$
Just as the name suggests.
