Manipulation of $C^1$ funcitons

Question

I think I read somewhere that composition of $C^1$ functions is also $C^1$, but I could not find the reference now. Also, is the difference of two $C^1$ functions still a $C^1$ function, please? And in general, what kind of algebraic manipulation of $C^1$ functions preserves the $C^1$ property, please? Thank you!

Answer 1

Chain Rule implies that the composition of $C^1$ functions is $C^1$. Note this applies to multivariable functions too. Consider $M(u,v)=uv$, the multiplication function, and $S(u,v)=u-v$, the subtraction function. These are $C^1$ smooth (they are polynomials after all). So, the composition of $x\mapsto (f(x),g(x))$ with $(u,v)\mapsto uv$ is also $C^1$; and this composition is nothing but the product $f(x)g(x)$.

When we get to division $D(u,v) = u/v$, things get interesting: this function is smooth only when $v\ne 0$. So, $f/g$ is guaranteed to be $C^1$ only in the region where $g\ne 0$. You can lose smoothness when dividing $C^1$ functions, even if the quotient is continuous: e.g., dividing $$f(x) = \begin{cases} x^3 \sin(1/x) \quad &x\ne 0\\ 0, \quad &x=0 \end{cases}$$ by $g(x)=x$ and extending to $0$ by continuity, we get $$f(x)/g(x) = \begin{cases} x^2 \sin(1/x) \quad &x\ne 0\\ 0, \quad &x=0 \end{cases}$$ which is not $C^1$.

Summary: the multivariable chain rule essentially contains all rules for algebraic manipulation of functions that preserve smoothness. Taking the inverse (which isn't an algebraic manipulation) stands aside from it; this is what the Inverse Function theorem is for.