Let $f$ be a real-valued function on an open subset $U$ of $\mathbb{R}$. Let $x_0 \in U$. We say that $f$ is differentiable at $x_0$ if $$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$$ exists. If it exists, this limit, often denoted $f'(x_0)$, is called the derivative of $f$ at $x_0$. If $f'(x_0)$ exists for all $x_0 \in U$ then $f$ is differentiable on $U$. The function $f'$, often denoted $df/dx$, is called the derivative of $f$.

Proposition: Basic Properties of Derivatives

Let $U$ be an open subset of $\mathbb{R}$, $f : U \to \mathbb{R}$. If $f$ is differentiable at $x_0 \in U$ then $f$ is continuous at $x_0$.
Let $f,g$ be real-valued functions on an open subset $U$ of $\mathbb{R}$. If $f$ and $g$ are differentiable at the point $x_0 \in U$, then so are $f+g$, $f - g$, $fg$, and, if $g(x_0) \neq 0$, $f/g$. Their derivatives at $x_0$ are given by the formulas $$(f+g)'(x_0) = f'(x_0) + g'(x_0)$$ $$(f - g)'(x_0) = f'(x_0) - g'(x_0)$$ $$(fg)'(x_0) = f(x_0)g'(x_0) + f'(x_0)g(x_0)$$ $$(f/g)'(x_0) = \frac{f'(x_0)g(x_0) - f(x_0)g'(x_0)}{(g(x_0))^2}$$
Let $U,V$ be open subsets of $\mathbb{R}$, and let $f: U \to V$, $g : V \to \mathbb{R}$ be functions. Let $x_0 \in U$ be such that $f'(x_0)$ and $g'(f(x_0))$ exist. Then $(g \circ f)'(x_0)$ exists and $$(g \circ f)'(x_0) = g'(f(x_0))f'(x_0)$$

Proofs:

We know that $f$ is differentiable at $x_0$, which is equivalent to saying that $$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}$$ exists. This equation is equivalent to saying that given any $\epsilon$, we have a $\delta$ such that $$| \frac{f(x) - f(x_0)}{x - x_0} - f'(x_0) | \leq \epsilon$$ whenever $x$ is within the $\delta$-ball around $x_0$. This equation is equivalent to: $$|f(x) - f(x_0) - f'(x_0)(x-x_0)| \leq \epsilon |x - x_0|$$ Now pick an $\epsilon_0$ and a suitable $\delta_0$ such that $$|f(x) - f(x_0) - f'(x_0)(x-x_0)| \leq \epsilon_0 |x-x_0|$$ whenever $x$ is within the $\delta_0$-ball around $x_0$. Thus we have that: $$|f(x) - f(x_0)| \leq |f(x) - f(x_0) - f'(x_0)(x-x_0)| + |f'(x_0)(x-x_0)|$$ $$\leq (\epsilon_0 + |f'(x_0)|)|x-x_0|$$ Now for any $\epsilon$ we can choose $\delta = \text{min}\{\delta_0,\epsilon/(\epsilon_0 + |f'(x_0)|)\}$, we have $|f(x)-f(x_0)| < \epsilon$ whenever $|x-x_0| < \delta$, proving continuity.
$$(f+g)'(x_0) = \lim_{x \to x_0} \frac{f(x) + g(x) - f(x_0) - g(x_0)}{x - x_0}$$ $$= f'(x_0) + g'(x_0)$$ Same as above for $(f-g)'(x_0)$. $$(fg)'(x_0) = \lim_{x \to x_0} \frac{f(x)g(x) - f(x_0)g(x_0)}{x - x_0}$$ $$= \lim_{x \to x_0} (f(x)\frac{g(x)-g(x_0)}{x-x_0} + g(x_0) \frac{f(x)-f(x_0)}{x-x_0})$$ $$ = \lim_{x \to x_0} f(x) \cdot \lim_{x \to x_0}f(x)\frac{g(x)-g(x_0)}{x-x_0} + g(x_0) + \lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}$$ $$ = f(x_0)g'(x_0) + g(x_0) f'(x_0)$$ To find $(f/g)'(x_0)$, it's easier to first find $(1/g)'(x_0)$ and use product rule $$(1/g)'(x_0) = \lim_{x \to x_0} \frac{(1/g(x)) - (1/g(x_0))}{x - x_0} = \lim_{x \to x_0} \frac{g(x_0) - g(x)}{(x-x_0)g(x_0) g(x)}$$ $$= -\frac{\lim_{x \to x_0}\frac{g(x) - g(x_0)}{x - x_0}}{g(x_0)\lim_{x \to x_0}(g(x))} = - \frac{g'(x_0)}{(g(x_0))^2}$$ We can now apply product rule to obtain required result.
For any fixed $y_0$ for which $g'(y_0)$ exists, set: $$A(y,y_0) := \begin{cases} \frac{g(y) - g(y_0)}{y - y_0} \ \ \ \text{ if } y \in V, y \neq y_0 \\ g'(y_0) \qquad\text{ if } y = y_0\end{cases}$$ Notice that we have $$g(y) - g(y_0) = A(y,y_0)(y - y_0)$$ for all $y \in V$. Also $$\lim_{y \to y_0} A(y,y_0) = g'(y_0) = A(y_0,y_0)$$ So $A$ is continuous at $y_0$. Setting $y_0 = f(x_0)$ and $y = f(x)$, we have that: $$(g \circ f)'(x_0) = \lim_{x \to x_0} \frac{g(f(x)) - g(f(x_0))}{x - x_0} = \lim_{x \to x_0} \frac{A(f(x),f(x_0))(f(x) - f(x_0))}{x-x_0}$$ $$= g'(f(x_0))f'(x_0)$$

Proposition: The Mean Value Theorem

Let $f$ be a real-valued function on an open subset $U$ of $\mathbb{R}$ that attains a maximum or a minimum at the point $x_0 \in U$. Then if $f$ is differentiable at $x_0$, $f'(x_0) = 0$.
Let $a,b \in \mathbb{R}$, $a < b$, and let $f$ be a continuous real-valued function on $[a,b]$ that is differentiable on $(a,b)$. Then there exists a number $c \in (a,b)$ such that $$f(b) - f(a) = (b-a)f'(c)$$

Proofs:

If $f'(x_0) \neq 0$, there exists a real number $\delta > 0$ such that if $x \neq x_0$ and $|x - x_0| < \delta$ then $$|\frac{f(x) - f(x_0)}{x - x_0} - f'(x_0) | < \frac{|f'(x_0)|}{2}$$ Thinking about it a bit, this is equivalent to $$f'(x_0) - \frac{|f'(x_0)|}{2} < \frac{f(x) - f(x_0)}{x - x_0} < f'(x_0) + \frac{|f'(x_0)|}{2}$$ Thinking about this a bit, we see that the middle section of this inequality must be always positive or always negative for all $x$ in a $\delta$-ball around $x_0$; but clearly this cannot be the case if $f(x_0)$ is to be a maximum or a minimum. Hence our proposition is true.
First we must prove a lemma, Rolle's Theorem, which states that if we have $a,b \in \mathbb{R}, a < b$ and $f$ a continuous real-valued function on $[a,b]$ that is differentiable on $(a,b)$ such that $f(a) = f(b) = 0$, then there must exist a number $c \in (a,b)$ such that $f'(c) = 0$. Because $[a,b]$ is compact, we can use Josh's lemma to show that there must exist a maximum and a minimum attained by $f$ on the compact set. If the maximum and the minimum is attained at the endpoints $a,b$, then trivially $f(x) = 0$ for $x \in [a,b]$ hence any $c$ will do the trick. If not, then we can use the previous proposition.

For an arbitrary function $f$, we can define another function $F$ such that $$F(x) = f(x) - f(a)- \frac{f(b) - f(a)}{b - a} \cdot (x - a)$$ $F$ can be visualised as the height of $f$ above a line connecting the points $f(a)$ and $f(b)$. By Rolle's Theorem, we have that for some $c$: $$F'(c) = f'(c) - \frac{f(b) - f(a)}{b-a} = 0$$ Proving the result.

Proposition: Taylor's Theorem

Let $U$ be an open interval in $\mathbb{R}$ and let the function $f : U \to \mathbb{R}$ be $(n+1)$ times differentiable. If for any $a,b \in U$ we defined $R_n(b,a) \in \mathbb{R}$ by the equation $$f(b) = f(a) + \frac{f'(a)(b-a)}{1!} + \frac{f''(a)(b-a)^2}{2!} + ...$$ $$+ \frac{f^{(n)}(a)(b-a)^n}{n!} + R_n(b,a)$$ then $$\frac{d}{dx} R_n(b,x) = -\frac{f^{(n+1)}(x)(b-x)^n}{n!}$$
Let $U$ be an open interval in $\mathbb{R}$ and let the function $f : U \to \mathbb{R}$ be $(n+1)$ times differentiable. Then for any $a,b \in U$ we have $$f(b) = f(a) + \frac{f'(a)(b-a)}{1!} + \frac{f''(a)(b-a)^2}{2!} + ...$$ $$ + \frac{f^{(n)}(a)(b-a)^n}{n!} + \frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1}$$ where $c$ is some number between $a$ and $b$ (or, if $a$ and $b$ are equal, $c = a$)

Proofs:

For any $x \in U$ we have $$f(b) = f(x) + f'(x)\frac{(b-x)}{1!} + f''(x)\frac{(b-x)^2}{2!} + ...$$ $$ + f^{(n)}(x)\frac{(b-x)^n}{n!} + R_n(b,x)$$ By differentiating both sides and cancelling like terms, we get the desired result.
This is trivial if $a = b$, so assume $a \neq b$. We need to show that $$R_n(b,a) = \frac{f^{(n+1)}(c)}{(n+1)!} (b-a)^{n+1}$$ for some $c$ between $a$ and $b$. There must exist some unique real number $K$ such that $$R_n(b,a) = K\frac{(b-a)^{n+1}}{(n+1)!} $$ Define a function $g : U \to \mathbb{R}$ such that: $$g(x) = R_n(b,x) - K\frac{(b-x)^{n+1}}{(n+1)!}$$ $g$ is differentiable on $U$, it also has the property that $g(a) = g(b) = 0$. Hence we can apply Rolle's Theorem, and get the result we wanted.