The Derivative at a Point

For the cart with position $s(t)=t^2$, the average velocity from $a$ to $a+h$ was $2a+h$. Both positive and negative increments approach the same value $2a$. We can now define the instantaneous velocity without assigning a meaning to division by zero: take the limit of the nonzero-interval average velocities.

For a function defined on an open interval containing $a$, the derivative at $a$ is the finite limit

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,$$

if it exists. An equivalent form uses $x=a+h$ and lets $x\to a$. Both formulations compare the output at the target with outputs at nearby inputs. We use a two-sided limit for interior differentiability. Endpoint derivatives, when useful, will be named one-sided derivatives.

A quadratic from the definition

Let $f(x)=2x^2-3x+1$. Expanding gives

$$\frac{f(a+h)-f(a)}h =\frac{4ah+2h^2-3h}h=4a+2h-3,\qquad h\ne0.$$

Therefore $f'(a)=4a-3$. At $a=2$, the function value is $3$ and derivative is $5$. The tangent line has equation

$$y=f(2)+f'(2)(x-2)=3+5(x-2).$$

A tangent is specified by its local limiting slope. It need not stay on one side of the curve or touch it at only one point. The tangent $y=0$ to $x^3$ at zero crosses the curve there.

Derivatives carry units

If $f$ measures temperature in degrees Celsius and $x$ measures hours, $f'(a)$ has units degrees Celsius per hour. A derivative of $-2$ means temperature is locally decreasing at that rate; it does not mean the temperature is negative. If position is measured in metres and time in seconds, velocity has units metres per second and acceleration, the derivative of velocity, has units metres per second squared.

For $s(t)=t^3-3t$ on a specified time interval, $v(t)=3t^2-3$. Positive velocity indicates motion in the positive coordinate direction. Speed is $|v(t)|$, so negative velocity is not negative speed. Later we will integrate speed to compute total distance.

Continuous does not mean differentiable

For $f(x)=|x|$ at zero, the difference quotient is $|h|/h$. It equals $-1$ for negative $h$ and $1$ for positive $h$. Thus the two-sided derivative does not exist even though $|x|\to0=f(0)$. The graph has a corner.

The function $x^{1/3}$ at zero has quotient $h^{-2/3}$, which grows without bound. This is often described geometrically as a vertical tangent. It is not a finite derivative under our definition. A discontinuity also prevents differentiability, but a smooth-looking numerical plot is not proof that discontinuities or corners are absent.

Why differentiability forces continuity

Suppose $f'(a)$ exists and is finite. For $h\ne0$,

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

As $h\to0$, the first factor tends to zero and the second tends to $f'(a)$. The product tends to zero. Hence $f(a+h)\to f(a)$, establishing continuity at $a$. Finiteness of the derivative matters because the ordinary product limit law needs finite limits.

This proof gives a useful diagnostic: if continuity fails, a finite derivative cannot exist. Its converse remains false, as the absolute-value example shows. A theorem and its converse are separate claims.

The local linear model

Rearrange the defining quotient into

$$f(a+h)=f(a)+f'(a)h+r(h),\qquad \frac{r(h)}h\longrightarrow0.$$

The derivative supplies a linear change, and the residual is small relative to the increment, not merely small in absolute size. For $f(x)=x^2$, $r(h)=h^2$, so $r(h)/h=h\to0$. This relative-error statement distinguishes a tangent approximation from any arbitrary line passing through the point.

The secant laboratory displays exact formulas for the square, cube, and absolute-value functions. A finite list of small increments cannot prove a derivative. The associated algebra explains the limit; the table and graph help us inspect its consequences. For polynomials the program evaluates algebraically simplified quotients to avoid catastrophic subtraction at small increments.

Exercises

1. Use the definition to differentiate $f(x)=3x^2+2x$ at a general $a$.
2. Find the tangent to $x^2$ at $x=-1$.
3. Use first principles to find the derivative of $1/x$ at $a\ne0$.
4. Show that the tangent to $x^3$ at zero crosses its graph.
5. A temperature derivative is zero at noon. Must the temperature be zero? Must it remain constant afterward?
6. Explain the different reasons that $|x|$ and $x^{1/3}$ lack a finite derivative at zero.

Worked solutions

1. Expansion gives $[f(a+h)-f(a)]/h=6a+3h+2$ for nonzero $h$, hence $f'(a)=6a+2$.
2. The point is $(-1,1)$ and slope $-2$, so $y=1-2(x+1)=-2x-1$.
3. The quotient simplifies to $-1/[a(a+h)]$, provided $h\ne0$ and $a+h\ne0$. Its limit is $-1/a^2$.
4. The derivative at zero is $\lim h^2=0$, so the tangent is the horizontal axis. Cubes have opposite signs on opposite sides of zero.
5. Neither. The derivative describes an instantaneous rate, not the level or a future interval. For example, $T(t)=20+t^2$ has derivative zero at $t=0$ without being constant.
6. Absolute value has two distinct finite one-sided slopes; the cube root has a quotient unbounded in the positive direction. Neither gives the required finite two-sided limit.