Derivatives Measure Local Change

An average rate compares two points. A derivative asks what the rate becomes when those points collapse into one. That small conceptual move gives mathematics a language for motion, growth, optimization, and sensitivity.

From secant to tangent

Between $x$ and $x+h$, the average rate of change of $f$ is

$$\frac{f(x+h)-f(x)}{h}.$$

The numerator measures output change; the denominator measures input change. To isolate the rate at $x$, we let the interval shrink:

For $f(x)=x^2$,

$$\frac{(x+h)^2-x^2}{h}=2x+h,$$

so $f'(x)=2x$. The slope depends on where we stand: at $x=3$, the local slope is $6$.

A derivative has units

If $s(t)$ is measured in meters and $t$ in seconds, then $s'(t)$ is measured in meters per second. If cost $C(q)$ is measured in dollars and quantity $q$ in items, then $C'(q)$ is dollars per item.

Units reveal the derivative’s meaning: it always measures output change per unit input change.

Local linearity

Near $x=a$, a differentiable function satisfies

$$f(a+\Delta x)\approx f(a)+f'(a)\Delta x.$$

This approximation predicts small changes without recalculating the entire function. It is also why derivatives power error estimates and numerical methods.

Differentiability is stronger than continuity

A function must be continuous wherever it is differentiable, but continuity alone is not enough. The function $f(x)=|x|$ is continuous at zero, yet its left slope is $-1$ and its right slope is $1$. The corner prevents a single derivative from existing.

Derive and interpret

For $f(x)=1/x$, the difference quotient simplifies to $-1/[x(x+h)]$, so $f'(x)=-1/x^2$. Its negative sign agrees with the graph’s decreasing behavior on each domain interval.

Check your understanding

Does $f'(a)=0$ guarantee an extremum?

Show the reasoning

No. $f(x)=x^3$ has $f'(0)=0$ but increases through zero.