Exercise 5.2.4

Question

Follow these steps to provide a slightly modified proof of the Chain Rule.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that a function $h: A ightarrow \mathbf{R}$ is differentiable at $a \in A$ if and only if there exists a function $l: A ightarrow \mathbf{R}$ which is continuous at $a$ and satisfies
    $$
    h(x)-h(a)=l(x)(x-a) \quad 	ext { for all } x \in A
    $$
  \item Use this criterion for differentiability (in both directions) to prove Theorem 5.2.5.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item First suppose $h$ is differentiable at $a$, then we can define
    $$
    l(x) = \begin{cases}
      \frac{h(x)-h(a)}{x-a} &{x 
e a} \
      h'(a) &{x = a}
    \end{cases}
    $$
    Since $\lim_{x	o a} l(x) = l(a)$, $l$ is continuous at $a$.

Now suppose $l : A 	o \mathbf{R}$ exists and satisfies $h(x) - h(a) = l(x)(x-a)$, dividing by $(x-a)$ gives
    $$
    \frac{h(x)-h(a)}{x-a} = l(x) \quad(x 
e a)
    $$
    Taking the limit of both sides as $x 	o a$ gives $h'(a) = l(a)$ (keep in mind the constraint $x 
e a$ makes no difference for the limit).
  \item Let $f$ be differentiable at $a$ and $g$ be differentiable at $f(a)$. we will show $g(f(x))$ is differentiable at $a$ with derivative $g'(f(a))f'(a)$.

Multiply the top and bottom by $f(y)-f(a)$ to get
    $$
    (g \circ f)'(a) = \lim_{y 	o a} \frac{g(f(y)) - g(f(a))}{f(y)-f(a)} \cdot \frac{f(y)-f(a)}{y-a}
    $$
    We're almost done, the right hand side is $f'(a)$ we just need to evaluate the nested limit on the left. Define $l(y) = \frac{g(y) - g(f(a))}{y - f(a)}$ and $l(a) = g'(f(a))$ then we have a product of continuous functions so we can use the algebraic limit theorem
    $$
    \lim_{y 	o a} l(f(y)) \cdot \frac{f(y)-f(a)}{y-a} = g'(f(a)) \cdot f'(a)
    $$
   \end{enumerate}

Exercise 5.2.4

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Exercise 5.2.4

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