Derivative based on continuity I have a question about whether I am even close to correct.
Let $\mathbb{I}$ and $\mathbb{J}$ be open intervals, and the functions $f:\mathbb{I} \to R$ and $h:\mathbb{J}\to R$ have the property that $h(\mathbb{J}) \subset \mathbb{I}$, so the composition $f\circ h :\mathbb{J} \to R$ is defined. Show that if the following conditions are true


*

*$x_o$ is in $\mathbb{J}$

*$h$ is continuous at $x_o$

*$h(x)\neq h(x_o)$ if $x \neq x_o$

*$f$ is differentiable at $h(x_o)$


then  
$$\lim_{x\to x_0} \frac{f(h(x)) - f(h(x_o))}{h(x) - h(x_o)} = f'(h(x_o)).$$
Proof I found:
Let $y = h(x)$ and $y_o = h(x_0)$. Then
$$\lim_{x\ \rightarrow x_o} \frac{f(h(x)) - f(h(x_o))}{h(x) - h(x_o)} =  \lim_{y\ \rightarrow y_o} \frac{f(y) - f(y_o)}{y - y_o}.$$
But by continuity of $h$, as $x$ $\rightarrow$ $x_o$, $h(x)$ $\rightarrow$ $h(x_o)$ or $y$ $\rightarrow$ $y_o$. Using this fact and the differentiability of $f$, we have
$$\lim_{y\ \rightarrow y_o} \frac{f(y) - f(y_o)}{y - y_o} = f’(y_o) = f’(h(x_o)).$$ 
$Q.E.D.$
This seems vastly too simple compared to the torrent of information given in the question, but if this proof works, that brings me to my second question related to the question immediately following this one. It says “Use exercise 6 to show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is differentiable at $x_o$ = 1, then...” and I will post a picture because I do not wish to retype it all with LaTeX. http://oi47.tinypic.com/2nb93zb.jpg
Basically, I struggled to understand the first question but when I finally did, the solution seemed trivial and I suspect this second question will be the same way, but I have no idea what they are asking for, nor do I see how I can use this first proof. I would greatly appreciate any insight.
 A: First, notice that your conditions are not that many. The first 3 statements translate to the following (less technical) statement - "The function $h(x)$ attains a unique value at $x_0 \in \mathbb{J}$". (This is a bit of tangent but it's useful to know how to translate between such technical statements and shorter, clearer statements).
Your proof is essentially correct but is missing a few details. I'll explain some of them and add some more information to help with your understanding.
First, what conditions do we need to use if the expression in the difference quotient is to even make sense? We use precisely the beginning of the statement of the problem - that $f\circ h$ is defined.
Next, as you let $x \to x_0$ you don't want to suddenly have to deal with a $h(x) - h(x_0) = 0$ and this is why you need the uniqueness of the value $h(x_0)$. This problem can be circumvented if in the calculation of the limit we work with an small open interval around $x_0$ that does not contain points $x_1$ such that $h(x_1) = h(x_0)$.
The continuity of $h(x)$ at $x = x_0$ ensures that $y = h(x) \to y_0 = h(x_0)$ as $x \to x_0$, and this is why you can write the limit as
$$\lim_{x\to x_0} \frac{f(h(x))-f(h(x_0))}{h(x)-h(x_0)} = \lim_{y\to y_0} \frac{f(y)-f(y_0)}{y-y_0}.$$
Finally, the differentiability of $f$ at $h(x_0)$ allows you to write the last equality. As you can see, there is a lot going on in the proof as well as in the statement of the problem. The algebraic part is a one-liner but the careful justifications take more time.
A: Your proof of Exercise 6 is correct, the point being (loosely) that the condition $y \to y_0, y \neq y_0$ includes $h(x) \to h(x_0), h(x) \neq h(x_0)$ as a 'special case'.
I find the $\epsilon$-$\delta$ approach a little clearer, if more tedious. If $f$ is differentiable at $y_0 = h(x_0)$, then for all $\epsilon>0$, there exists $\delta >0$ such that if $0 < |y-y_0| < \delta$, then $|\frac{f(y)-f(y_0)}{y-y_0} - f'(y_0)| < \epsilon$. Since $h$ is continuous at $x_0$, and $h(x) \neq h(x_0)=y_0$ for $x\neq x_0$ near $x_0$, there exists $\eta>0$ such that if $0 < |x-x_0| < \eta$, then $0 < |h(x)-h(x_0)| < \delta$, and hence $|\frac{f(h(x))-f(h(x_0))}{h(x)-h(x_0)} - f'(y_0)| < \epsilon$. The limit follows.
For Exercise 7, you need to find the the appropriate $h$ for (a)-(e). The last two parts involve a small twist. I will illustrate by solving (c) & (d), you will be able to do the others.
For (c), take $h(x) = x^2$, and $x_0 = 1$. If $x \neq 1$ and $|x-1| < 1$ (or at least on a sufficiently small open interval containing $1$), then $h(x) \neq h(x_0)$, and since $h$ is continuous, the conditions of Exercise 6 are satisfied. Consequently, the limit is $f'(h(x_0)) = f'(1)$.
For (d), we write the limit quotient  as $\frac{f(x^2)-f(1)}{x-1} =  \frac{f(x^2)-f(1)}{x^2-1} \frac{x^2-1}{x-1}$. If we take $h(x) = x^2$ and $x_0 = 1$, then we know the limit of the first fraction is $f'(1)$, and since the function $x \mapsto x^2$ is differentiable at $x=1$ with derivative $2$, we know the limit of the second fraction is $2$. Since multiplication is continuous, we know that the limit of the product is the product of the limits, so we have $\lim_{x \to 1} \frac{f(x^2)-f(1)}{x^2-1} \frac{x^2-1}{x-1} = \lim_{x \to 1} \frac{f(x^2)-f(1)}{x^2-1} \lim_{x \to 1} \frac{x^2-1}{x-1} = f'(1) 2$, as required.
