Differentiation

The derivative of \( f\colon A \subset R \to B : x\mapsto f(x) \) is defined as:

\[ f'(x) = \lim_{h \to 0} \left(\frac{f(x + h) - f(x)}{(x + h) - x}\right) \]

Note that \( f'(x) \) may not exist \( \forall x \in A \) even for continuous functions.

We will now prove some important identities using this definition. \pagebreak

The product rule is defined: \[ \frac{d}{dx}f(x)g(x) = f(x)g'(x) + f'(x)g(x) \]

Proof:

\begin{align} \frac{d}{dx}f(x)g(x) &= \lim_{h \to 0}(\frac{f(x + h)g(x + h) - f(x)g(x)}{(x + h) - x}) \nonumber \\ &= \lim_{h \to 0}(\frac{f(x + h)g(x + h) - f(x)g(x)}{h}) \nonumber \\ &= \lim_{h \to 0}(\frac{f(x + h)g(x + h) - f(x)g(x) + f(x)g(x + h) - f(x)g(x + h)}{h}) \nonumber \\ &= \lim_{h \to 0}(\frac{f(x)(g(x + h) - g(x)) + f(x + h)g(x + h) - f(x)g(x + h)}{h}) \nonumber \\ &= \lim_{h \to 0}(\frac{f(x)(g(x + h) - g(x))}{h} + g(x + h)\frac{f(x + h) - f(x)}{h}) \nonumber \\ &= f(x)g'(x) + f'(x)g(x) \nonumber \end{align}

Alternatively, set \( y = f(x)g(x) \), take logarithms and then implicitly differentiate. This proof requires use of the chain rule however.

The chain rule is defined: \[ \frac{d}{dx}f(g(x)) = g'(x)f'(g(x)) \]

Proof:

\begin{align} \frac{d}{dx}f(g(x)) &= \lim_{h \to 0}(\frac{f(g(x + h)) - f(g(x))}{(x + h) - x}) \nonumber \\ &= \lim_{h \to 0}(\frac{f(g(x + h)) - f(g(x))}{h}) \nonumber \\ &= \lim_{h \to 0}(\frac{f(g(x + h)) - f(g(x))}{g(x + h) - g(x)} \frac{g(x + h) - g(x)}{h}) \nonumber \\ \end{align}