Verify: A differentiable function has a continuous approximation function A function $s_a$ is defined as a secant function of $f$ at $a$ if $f(x) - f(a) = s_a(x)(x-a)$.  Prove that $f$ is differentiable at $a$ if and only if there exists a secant function $s_a$ continuous at $a$, in which case $s_a(a) = f'(a)$.  Use this to prove the Chain Rule.
Note: Proofs are available.  This question asks for verification and critique of this proof.
Proof: For a fixed $f$ and $a$, each value at $a$ uniquely determines exactly one secant function $s_a$.  Thus, a secant function continuous at $a$ exists if and only if there exists a value at $a$ which makes the corresponding $s_a$ continuous at $a$.  This, in turn, is met if and only if $s_a(x)$ has a limit as $x \to a$.  But
$$\lim_{x \to a}s_a(x) = \lim_{x \to a}\frac{f(x) - f(a)}{x-a} = f'(a).$$
Thus, if and only if $f'(a)$ exists, then a $s_a$ continuous at $a$ exists with $s_a(a) = f'(a)$.
Chain Rule: Given $f$ differentiable at $a$ and $g$ differentiable at $f(a)$, then $g \circ f$ is differentiable at $a$ with derivative $g'(f(a)) \cdot f'(a)$.
Proof: By the theorem above, we have
$$\begin{align*}f(x) - f(a) &= s_a(x)(x-a) \\
g(x) - g(f(a)) &= r_{fa}(x)(x - f(a))\end{align*}$$
with $s_a$ continuous at $a$ and $r_{fa}$ at ${f(a)}$. Consequently, $$q(x) := g'(f(x)) \cdot f'(x) = r_{fa}(f(x)) \cdot s_a(x)$$ is continuous at $x=a$.  Therefore
$$\begin{align*}q(x)(x-a) &= r_{fa}(f(x)) \cdot s_a(x)(x - a) \\
&= r_{fa}(f(x)) \cdot (f(x) - f(a))) \\
&= g(f(x)) - g({f(a)}) \\
&= (g \circ f)(x) - (g \circ f)(a)\\
\end{align*}$$
and $g \circ f$ is differentiable at $a$ with derivative $q(a)$, completing the proof.
Questions: Is the proof correct and rigorous? How could the writing be made more clear?

Update: I haven't received any responses.  Could someone comment if:

*

*The proof is correct (and so no comment because no obvious errors)

*The proof is incorrect, unclear, or otherwise jumbled (and so no comment because too jumbled to verify pr discuss)

 A: Your proofs are correct and rigorous basically, as I have checked. Here are some comments.
On the proof for the equivalence of the differentiability of $f$ and the existence of a continuous "secant function"
It is clearer to prove the "if" direction and "only if" direction separately in general. (There are exceptions such as transformations of equalities, which only support the general rule.) Nevertherless, your proof shows the equivalence very clearly.
On the proof for $g\circ f$ is differentiable
The definition, "$q(x) := g'(f(x)) \cdot f'(x)$" is wrong. $q(x)= g'(f(x)) \cdot f'(x)$ is only true when $x=a$ in general. What you wanted to say is simply "$q(x):=r_{fa}(f(x)) \cdot s_a(x)$".  Taking $x=a$, we know $q(a)=r_{fa}(f(a)) \cdot s_a(a)=g'(f(a))\cdot f'(a)$.

The following part is not about your proofs.
On the definition of "secant function"

A function $s_a$ is defined as a secant function of $f$ at $a$ if $f(x) - f(a) = s_a(x)(x-a)$.

There are multiple issues with this definition.

*

*A "secant function" has been universally defined as the reciprocal of the cosine function basically. It is confusing to assign a different meaning.

* The function maps an $x\not=a$ to the slope of the secant that cuts the graph of $f$ at point $(a, f(a))$ and $(x, f(x))$. I would call it "a slope function of $f$ relative to $a$" or "a slope function of $f$ about $a$".


*The domain and codomain of the function is not defined explicitly. This might be good for faster understanding so as to concentrate on the main content. However, it is not ideal for correctness and rigor.

*

*$f$ should be defined on a neighborhood of $a$.

*Also, $a$ should in the domain of $f$.

*The domain of $s_a$ can be defined the same as that of $f$.

*The codomain of $s_a$ can be the same as that of $f$, either real numbers or complex numbers.



