Prob. 2, Chap. 5 in Baby Rudin: How is the inverse function differentiable? Here is Prob. 2, Chap. 5 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

Suppose $f^\prime(x) > 0$ in $(a, b)$. Prove that $f$ is strictly increasing in $(a, b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that 
  $$ g^\prime\left( f(x) \right) = \frac{1}{f^\prime(x)} \qquad \qquad \qquad (a < x < b). $$

My effort: 

Let $x_1$ and $x_2$ be any two real numbers such that 
  $$ a < x_1 < x_2 < b.$$ 
  Then $f$ is differentiable on the interval $\left[ x_1, x_2 \right]$ and hence on the segment $\left( x_1, x_2 \right)$, and $f$ is of course continuous on the interval $\left[ x_1, x_2 \right]$. So, by the Mean-Value Theorem, we can find a point $x \in \left( x_1, x_2 \right)$, such that 
  $$ f\left( x_2 \right) - f\left( x_1 \right) = \left( x_2 - x_1 \right) f^\prime (x) > 0, $$
  and so 
  $$  f\left( x_1 \right) <  f\left( x_2 \right) \qquad \qquad \qquad ( a< x_1 < x_2 < b), $$
  and therefore $f$ is strictly increasing on $(a, b)$. 
As $f$ is strictly increasing on $(a, b)$, so it is injective, and therefore the inverse function $g$ exists; this $g$ is a mapping of $\mathrm{range} f$ into (rather onto) $(a, b)$, and is defined by 
  $$ g \left(  y \right) = x \ \mbox{ for all } \ y = f(x) \in \mathrm{range} f.$$ 
  Thus the mapping $h$ of $(a, b)$ into $(a, b)$, given by 
  $$ h(x) = g \left( f(x) \right) \qquad \qquad \qquad ( a< x < b),$$
  is the identity mapping. So 
  $$h^\prime(x) = 1 \qquad \qquad \qquad (a < x < b). \tag{1} $$
  But by Theorem 5.5 in Baby Rudin (i.e. the Chain Rule), we know that if $g^\prime$ exists at each point in the range of $f$, then we obtain 
  $$h^\prime(x) = g^\prime\left( f(x) \right) f^\prime(x) \qquad \qquad \qquad (a < x < b). \tag{2} $$
  From (1) and (2), we can conclude that 
  $$ g^\prime\left( f(x) \right) f^\prime(x) = 1 \qquad \qquad \qquad (a < x < b), $$
  and since $f^\prime(x) > 0$ for all $x \in (a, b)$, therefore 
  $$ g^\prime\left( f(x) \right) = \frac{1}{f^\prime(x)  } \qquad \qquad \qquad (a < x < b). $$

Is my reasoning so far correct? If so, then how to show that the function $g$ is indeed differentiable (at each point in the range of $f$)? 
Or, is there a problem in my reasoning above?
P.S.: 
After reading the above answers, I have arrived at this solution to my original problem and would be grateful for the appraisal thereof of the Math SE community. 

As $f$ is differentiable at each point in $(a, b)$, so $f$ is continuous in $(a, b)$, and as $f^\prime(x) > 0$ for every $x \in (a, b)$, so $f$ is strictly increasing on $(a, b)$. Thus $f$ is a continuous bijective mapping of $(a, b)$ onto the range of $f$. We now show that the range of $f$ is also an open interval. 
Suppose $y_1$ and $y_2$ are any two points in the range of $f$ such that $y_1 < y_2$, and suppose that $y$ is any real number such that $y_1 < y < y_2$. As $y_1$ and $y_2$ are in the range of $f$, so there exist some points $x_1$ and $x_2$ in $(a, b)$ such that $f\left(x_1 \right) = y_1$ and $f \left( x_2 \right) = y_2$; moreover, as $y_1 < y_2$ and as $f$ is strictly increasing, so we must also have $x_1 < x_2$, for otherwise we would obtain $f\left( x_1 \right) \geq f\left( x_2 \right)$. 
Now as $f$ is continuous on the interval $\left[ x_1, x_2 \right]$ and as $$y_1 = f \left( x_1 \right) < y < f \left( x_2 \right) = y_2,$$ so by the intermediate-value theorem for continuous functions, there is some point $x \in \left( x_1, x_2 \right)$ such that $y = f(x)$. 
Thus we have shown that, for any points $y_1$ and $y_2$ in the range of $f$ such that $y_1 < y_2$, the segment $\left( y_1, y_2 \right)$ is also contained in the range of $f$. That is, the range of $f$ is an interval. 
We now show that the range of $f$ is an open interval. For any point $y_0$ in the range of $f$, we have a unique point $x_0 \in (a, b)$ such that $y_0 = f\left(x_0\right)$. But as $a < x_0 < b$, so we also have $$ a < \frac{a+x_0}{2} < x_0 < \frac{x_0+b}{2} < b,$$ and therefore $$ f \left( \frac{a+x_0}{2} \right) < f\left( x_0 \right) < f \left( \frac{x_0+b}{2} \right);$$ that is, for any point $y_0$ in the range of $f$, we have points $f\left(\frac{a+x_0}{2} \right)$ and $f \left( \frac{x_0+b}{2} \right)$ in the range of $f$ such that $$ f\left(\frac{a+x_0}{2} \right) < y_0 < f \left( \frac{x_0+b}{2} \right).$$ So the range of $f$ has no maximum element and no minimum element. 
Hence the range of $f$ is a possibly infinite (on either side) open interval, say $(c, d)$. 
Let the function $g \colon (c, d) \to (a, b)$ be the inverse of the function $f \colon (a, b) \to (c, d)$. We show that $g$ is continuous by showing that the inverse image under $g$ of every open set in $(a, b)$ is open in $(c, d)$. For this it suffices to show that the inverse image under $g$ of every open interval $(u, v) \subset (a, b)$ is open in $(c, d)$. 
But as $f$ and $g$ are bijective and are the inverses of each other, so $$g^{-1} \left[ (u, v) \right] = f \left[ (u, v) \right]. $$
  We now show that $$f \left[ (u, v) \right] = \left( f(u), f(v) \right).$$ 
  Suppose $y \in f \left[ (u, v) \right]$. Then $y = f(x)$ for a (unique) point $x \in (u, v)$. As $a < u < x < v < b$ and as $f$ is strictly increasing on $(a, b)$, so we must have $f(u) < f(x) < f(v)$, that is, $y \in \left( f(u), f(v) \right)$. 
Conversely, suppose that $y \in \left( f(u), f(v) \right)$. Then $f(u) < y < f(v)$, and $f$ is continuous on the closed interval $[u, v]$; so (by the intermediate-value theorem for continuous functions)  there is some (unique) point $x \in (u, v)$ such that $y = f(x)$, which implies that $y \in f\left[ (u, v) \right]$. 
Therefore, $f \left[ (u, v) \right] = \left( f(u), f(v) \right)$, which is open in $(c, d)$, as required. 
Thus the inverse function $g \colon (c, d) \to (a, b)$ is continuous whenever $f^\prime(x) > 0$ for all $x \in (a, b)$. 
Now let $y_0 \in (c, d)$. Then there is a unique point $x_0 \in (a, b)$ such that $y_0 = f\left( x_0 \right)$, which implies that $g \left( y_0 \right) = x_0$. 
Since $g$ is continuous at $y_0$, so $$ \lim_{y \to y_0} g(y) = g\left( y_0 \right); $$ 
  that is, $$\lim_{y \to y_0} x = x_0,  \ \mbox{ where } \ x = g(y); $$
  thus, as $y \to y_0$ in $(c, d)$, the point $x = g(y) \to x_0$ in $(a, b)$. 
Now as $f^\prime\left( x_0 \right) \neq 0$, so we find that 
  $$g^\prime \left( y_0 \right) = \lim_{y \to y_0 } \frac{ g(y) - g\left( y_0 \right) }{ y - y_0 } = \lim_{x \to x_0 }  \frac{ x - x_0 }{ f(x) - f \left( x_0 \right) } = \lim_{x \to x_0 } \frac{ 1 }{ \frac{ f(x) - f\left( x_0 \right) }{ x - x_0 } } = \frac{1}{ f^\prime \left( x_0 \right) },$$ 
  which shows that $g^\prime$ exists at each point $y_0 \in (c, d)$, and also that $$ g^\prime \left( y_0 \right) = \frac{1}{ f^\prime \left( x_0 \right) },$$ as required. 

Have I finally managed to get this solution correct? 
 A: The problem in getting $g'(f(x))=1/f'(x)$ from $g(f(x))=x$ is exactly that you have first to prove that $g$ is differentiable.

Lemma. If $f$ is strictly increasing and continuous on $(a,b)$, then the range of $f$ is an open interval and the inverse function of $f$ is continuous.

You can try your hand in proving the lemma. With it, the proof of the inverse function theorem on derivatives is easy. Let $y_0$ belong to the range of $f$, with $y_0=f(x_0)$; then
$$
\lim_{y\to y_0}\frac{g(y)-g(y_0)}{y-y_0}=
\lim_{x\to x_0}\frac{x-x_0}{f(x)-f(y_0)}
$$
by using the substitution $x=g(y)$ in the limit, which is possible because $g$ is continuous.

A sketch of the proof of the lemma.
Assume $f$ is strictly increasing and continuous on $(a,b)$. Then, by the intermediate value theorem its range is an interval, which cannot contain its end points (if bounded). Indeed, suppose the range is upper bounded and let $s$ be the supremum. If $s=f(x)$, for some $x\in(a,b)$, then there exists $y\in(x,b)$ and $s=f(x)<f(y)$: a contradiction. Similarly for the infimum.
Let $g$ be the inverse function of $f$ (it exists because $f$ is injective). Then $g$ is increasing as well. Consider $c$ in the domain $I$ of $g$ (which is an open interval). Then it is easy to prove that
$$
\lim_{x\to c^-}g(x)=\sup\{g(t):t\in I, t<c\}
$$
The supremum exists, because $g(c)$ is an upper bound.
Similarly,
$$
\lim_{x\to c^+}g(x)=\inf\{g(t):t\in I, t>c\}
$$
and $g(c)$ is a lower bound. Now prove that both limits must be equal to $g(c)$ (hint: both limits belong to $(a,b)$), so $g$ is continuous.
A: Fix $y$ in the range of $f$. To show that $g'(y)$ exists we need to show that $$\lim_{h\to0}\frac{g(y+h)-g(y)}{h}$$ exists. For $h$ sufficiently small so that $y+h$ is still in the range of $f$, we can use the fact that $f$ is strictly increasing to obtain unique numbers $x_{y+h}$ and $x_y$ in the domain of $f$ such that $f(x_{y+h})=y+h$ and $f(x_y)=y$. The limit we need to show exists is now $$\lim_{h\to0}\frac{g(f(x_{y+h}))-g(f(x_y))}{h},$$ or since $g$ is the inverse of $f$, $$\lim_{h\to0}\frac{x_{y+h}-x_y}{h}$$ needs to be shown to exist. Write $$\frac{x_{y+h}-x_y}{h}=\frac{x_{y+h}-x_y}{f(x_{y+h})-f(x_y)}\frac{f(x_{y+h})-f(x_y)}{h}.$$ The right hand side is a product whose individual limits exist and are equal to $1/f'(x_y)$ and $1$ respectively. Therefore $$\lim_{h\to0}\frac{x_{y+h}-x_y}{h}=\frac{1}{f'(x_y)}.$$
