A twist to the definition of a derivative

Question

Two questions:

1. Functions are connections between numbers. Derivatives are connections between functions. So is there something like connections between derivatives?
2. What function does the following limit yield: $$\lim_{x \rightarrow a} \frac{ f' (x)- f' (a)}{f(x)-f(a)}$$ (for example if we plug in $\sin(x),$ then we get $- \tan(x),$ and if we plug in $x^{2},$ then we get $\frac{1}{x} $)

Answer 1

Let $f$ be a twice differentiable function such that $$Y(a)=\lim_{x\to a}\frac{f'(x)-f'(a)}{f(x)-f(a)}$$ exists and is a real number. Since $f$ is differentiable at $a$, we find that \begin{eqnarray} Y(a)\cdot f'(a) &=& \lim_{x\to a}\frac{f'(x)-f'(a)}{f(x)-f(a)}\cdot \lim_{x\to a}\frac{f(x)-f(a)}{x-a}\\ &=& \lim_{x\to a}\frac{f'(x)-f'(a)}{x-a}\\ &=& f''(a). \end{eqnarray} So for twice differentiable functions such that $Y(a)$ exists and such that $f'(a)\neq 0$, we find that $Y=\frac{f''(a)}{f'(a)}$. This gives a nice answer as to what this limit actually is for this class of functions. Alternatively, you can use l'Hôpitals rule to calculate this limit as suggested by Ultralegend5385 in the comments.

Now note that if $f(x)=c$ is a constant function, then $Y(a)=\lim_{x\to a}\frac{0}{f(x)-f(a)}=\lim_{x\to a}\frac{0}{0}$ does not exists as we cannot divide by zero. So the requirement that $Y(a)$ exists is not satisfied for all twice differentiable functions. It becomes even trickier for functions like $$f_n(x)=\begin{cases}x^n\sin(\frac{1}{x}) & \mbox{ if }x\neq 0,\\ 0 & \mbox{ if } x=0\end{cases}$$ to figure out whether $Y(0)$ even exists. All in all, I'd say it's not the most interesting limit to look at.