**Step 1: Sum rule**

By definition of the derivative:

\begin{align*} \frac{(f+g)(a+h) - (f+g)(a)}{h} = \frac{f(a+h) - f(a)}{h} + \frac{g(a+h) - g(a)}{h} \end{align*}

Since $f'(a)$ and $g'(a)$ both exist, the algebra of limits (the limit of a sum equals the sum of the limits, provided both limits exist) gives:

\begin{align*} (f+g)'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} + \lim_{h \to 0} \frac{g(a+h) - g(a)}{h} = f'(a) + g'(a) \end{align*}

**Step 2: Product rule**

Write:

\begin{align*} \frac{f(a+h)g(a+h) - f(a)g(a)}{h} &= \frac{f(a+h)g(a+h) - f(a)g(a+h) + f(a)g(a+h) - f(a)g(a)}{h} \\ &= \frac{f(a+h) - f(a)}{h} \cdot g(a+h) + f(a) \cdot \frac{g(a+h) - g(a)}{h} \end{align*}

Since $g$ is differentiable at $a$, it is continuous at $a$ (differentiability implies continuity), so $\lim_{h \to 0} g(a+h) = g(a)$. By the algebra of limits (the limit of a product equals the product of the limits, provided both limits exist):

\begin{align*} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \cdot g(a+h) = f'(a) \cdot g(a) \end{align*}

and $\lim_{h \to 0} f(a) \cdot \frac{g(a+h) - g(a)}{h} = f(a) \cdot g'(a)$. Therefore $(fg)'(a) = f'(a) g(a) + f(a) g'(a)$.

**Step 3: Quotient rule**

First show that $1/g$ is differentiable at $a$ when $g(a) \neq 0$. Since $g$ is differentiable at $a$, it is continuous at $a$. Since $g(a) \neq 0$ and $g$ is continuous, there exists $\delta > 0$ such that $g(a+h) \neq 0$ for $|h| < \delta$. For such $h$:

\begin{align*} \frac{\frac{1}{g(a+h)} - \frac{1}{g(a)}}{h} &= \frac{g(a) - g(a+h)}{h \cdot g(a+h) g(a)} = -\frac{1}{g(a+h) g(a)} \cdot \frac{g(a+h) - g(a)}{h} \end{align*}

As $h \to 0$, $g(a+h) \to g(a)$ (by continuity) and $\frac{g(a+h) - g(a)}{h} \to g'(a)$. By the algebra of limits:

\begin{align*} \left(\frac{1}{g}\right)'(a) = -\frac{1}{g(a) \cdot g(a)} \cdot g'(a) = -\frac{g'(a)}{g(a)^2} \end{align*}

Applying the product rule from Step 2 to $f \cdot (1/g)$:

\begin{align*} \left(\frac{f}{g}\right)'(a) = f'(a) \cdot \frac{1}{g(a)} + f(a) \cdot \left(-\frac{g'(a)}{g(a)^2}\right) = \frac{f'(a)g(a) - f(a)g'(a)}{g(a)^2} \end{align*}

**Step 4: Chain rule**

Define the error function $\phi: E \to \mathbb{R}$ by:

\begin{align*}

\phi(y) = \begin{cases} \frac{f(y) - f(g(b))}{y - g(b)} - f'(g(b)) & \text{if } y \neq g(b) \\ 0 & \text{if } y = g(b) \end{cases} \end{align*} Since $f$ is differentiable at $g(b)$, the limit $\lim_{y \to g(b)} \frac{f(y) - f(g(b))}{y - g(b)} = f'(g(b))$ exists. Therefore $\lim_{y \to g(b)} \phi(y) = 0 = \phi(g(b))$, so $\phi$ is continuous at $g(b)$. For all $y$ (including $y = g(b)$, where both sides equal zero): \begin{align*} f(y) - f(g(b)) = (y - g(b))(f'(g(b)) + \phi(y)) \end{align*} Setting $y = g(b+h)$: \begin{align*} \frac{f(g(b+h)) - f(g(b))}{h} = \frac{g(b+h) - g(b)}{h} \cdot (f'(g(b)) + \phi(g(b+h))) \end{align*} As $h \to 0$: the first factor tends to $g'(b)$ (by differentiability of $g$ at $b$). For the second factor: since $g$ is differentiable at $b$ it is continuous at $b$, so $g(b+h) \to g(b)$, and since $\phi$ is continuous at $g(b)$, $\phi(g(b+h)) \to \phi(g(b)) = 0$. By the algebra of limits: \begin{align*} (f \circ g)'(b) = g'(b) \cdot (f'(g(b)) + 0) = g'(b) \cdot f'(g(b)) \end{align*}