In Calculus 1 the derivative was the limit of difference quotients, and the rules took over quickly. In analysis we go back to the definition, now with ε–δ limits behind it, because the interesting functions are exactly the ones the rules cannot handle: pieces glued together, or functions that oscillate near a point.
Idea clave
$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ when this limit exists (equivalently $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$). To use it: write the difference quotient, simplify it algebraically for $h \ne 0$, then take the limit, often with the squeeze theorem.
Por qué funciona
The difference quotient is the slope of the secant line through $(a, f(a))$ and $(a + h, f(a + h))$. As $h \to 0$ the secant turns into the tangent, and the derivative is its slope. Because the definition is a limit, every tool for limits applies: algebra to cancel the $h$, the squeeze theorem for oscillating factors, and one-sided limits at a corner.
Worked examples
Ejemplo resuelto: A polynomial from scratch
For $f(x) = 2x^{2} + 3x$, simplify $\frac{f(1 + h) - f(1)}{h}$ and find $f'(1)$.
- $f(1 + h) = 2(1 + 2h + h^{2}) + 3 + 3h = 5 + 7h + 2h^{2}$, and $f(1) = 5$.
- So the quotient is $\frac{7h + 2h^{2}}{h} = 7 + 2h$ for $h \ne 0$.
Respuesta: $7 + 2h \to 7$, so $f'(1) = 7$.
Ejemplo resuelto: An oscillating function that is differentiable
Let $f(x) = x^{2}\sin\frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$. Find $f'(0)$.
- $\frac{f(h) - f(0)}{h} = h\sin\frac{1}{h}$.
- $\left|h\sin\frac{1}{h}\right| \le |h| \to 0$, so by the squeeze theorem the limit is $0$.
Respuesta: $f'(0) = 0$, even though $f'(x) = 2x\sin\frac{1}{x} - \cos\frac{1}{x}$ has no limit as $x \to 0$.
Ejemplo resuelto: One power less, and it fails
Let $g(x) = x\sin\frac{1}{x}$ for $x \ne 0$ and $g(0) = 0$. Does $g'(0)$ exist?
- $\frac{g(h) - g(0)}{h} = \sin\frac{1}{h}$.
- Along $h = \frac{1}{2\pi n}$ it is $0$; along $h = \frac{1}{2\pi n + \pi/2}$ it is $1$.
Respuesta: No limit, so $g'(0)$ does not exist (although $g$ is continuous at $0$).
Error común
Using the rules at a break point. For $f(x) = x^{2}$ ($x \le 1$) and $f(x) = 3x - 2$ ($x > 1$), "the derivative is $2x$ on the left and $3$ on the right" describes $f'$ away from $1$, but at $x = 1$ only the definition decides: the left quotient tends to $2$ and the right one to $3$, so $f'(1)$ does not exist. Likewise for $x^{2}\sin\frac{1}{x}$: the formula for $f'(x)$ at $x \ne 0$ says nothing about $f'(0)$.