Difference Quotient and h-Method

The difference quotient $\frac{(f(x_{0}+h) - f(x_{0}))}{h}$ is ordinary fraction arithmetic: the slope of the secant through two curve points, i.e. the average rate of change over an interval. The differential quotient is its limit for h → 0: the two points move together, the secant tilts into the tangent, and out comes the local rate of change f′(x₀) at a single point. Mnemonic: difference = two points, differential = one point with a limit. Example $f(x) = x^{2}$ at $x_{0} = 3$: the difference quotient is 6 + h (still depends on h), the differential quotient is the number 6. Without the limit there is no derivative.

Four steps using $f(x) = x^{2}$, $x_{0} = 3$. First set up: $\frac{(f(3+h) - f(3))}{h} =\frac{ ((3+h)^{2} - 9)}{h}$. Second expand: $\frac{(9 + 6h + h^{2} - 9)}{h} =\frac{ (6h + h^{2})}{h}$. Third factor out h and cancel: $\frac{h\cdot{}(6 + h)}{h} = 6 + h$; cancelling is allowed because $h \ne{} 0$. Fourth take the limit: for h → 0, f′$(3) = 6$ remains. The critical point is step 2: only when every summand in the numerator contains an h can you cancel; if an h-free term remains, there is an arithmetic error in the binomial expansion. The method works the same for x³ (with (a+b)³) or 1/x (with a common denominator).

Substituting $h = 0$ directly gives $\frac{(f(x_{0}) - f(x_{0}))}{0} =\frac{ 0}{0}$, an indeterminate expression; division by 0 is undefined. Moreover the difference quotient describes a secant through two points, and at $h = 0$ both would coincide: no unique line passes through a single point. The h-method avoids this cleanly: for $h \ne{} 0$ the fraction is simplified algebraically until h is cancelled; the resulting term (say 6 + h) can then be evaluated at $h = 0$ too, and the limit exists. Exactly this two-step (cancel first, then h → 0) is the mathematical substance of the limit concept behind the derivative.

The average rate of change on an interval [a; b] is the difference quotient $m =\frac{ (f(b) - f(a))}{(b - a)}$: total change divided by interval length, geometrically the secant slope. Example: $f(x) = x^{2}$ on [1; 3]: $m =\frac{ (9 - 1)}{2} = 4$. In applied contexts it carries a unit, e.g. metres per second (average speed) or degrees per hour. The local rate of change f′(x₀), by contrast, is the limiting case at one point, in the speedometer picture: average speed of the trip versus instantaneous speed at the moment of a speed check. A classic exam task is comparing the two values, here: average rate 4 on [1; 3], local rate f′$(2) = 4$ at the interval midpoint.

The derivative exists only if the limit of the difference quotient exists, from both sides with the same value. Three typical counterexamples: first, kinks like $f(x) = |x|$ at $x = 0$, where the left-hand secant slope gives −1 and the right-hand one +1; no common limit, no tangent. Second, jump points: if f is discontinuous, there can be no derivative at all, differentiability requires continuity. Third, vertical tangents as for $f(x) = \sqrt[3]{x}$ at 0, where the secant slopes grow towards infinity. Conversely, continuity does not imply differentiability; |x| is the standard example of a continuous function that is not differentiable everywhere.