What does a slope at a single point mean?

Start with the smooth cubic and drag $Q$ toward $P$. The orange secant pivots and its slope settles near the dashed tangent. Then choose Corner |x|: the secant slope is $1$ from the right and $-1$ from the left, however small $h$ becomes. Finally type $h=0$: the stage refuses to draw a secant, because two equal points determine no line.

Intuition: zooming in on a curve

A secant is the straight line through two points of a graph, $P=(x,f(x))$ and $Q=(x+h,f(x+h))$. Its slope is an ordinary average rate of change: rise over run. When $h$ is small, $Q$ is close to $P$ and the secant hugs the graph near $P$. For a smooth function the secant slopes approach one number. That number is the slope of the graph at $P$.

The important word is approach. The secant formula is never evaluated with $h=0$; the tangent slope is what the secant slopes tend to as $h$ shrinks through nonzero values.

Definition

The difference quotient of $f$ at $x$ with step $h\ne0$ is

The function $f$ is differentiable at $x$ when $f(x)$ exists and the two-sided limit

exists as a finite real number. The tangent line at $P$ is then $y=f(x)+f'(x)(t-x)$. The one-sided limits $h\to0^-$ and $h\to0^+$ are the left and right slopes; the derivative exists exactly when both exist, are finite and agree.

Mechanics: what the explorer computes

Each secant slope is an actual quotient of two function values with a nonzero step. For the cubic $f(x)=x^3/6-x$ and decimal inputs, the values are exact rationals, so the inspector shows $=$. For $\sin x$ they are binary64 approximations, marked $\approx$. The tangent slope row is not read from the table: it comes from the function’s analytic description (for a polynomial, the exact derivative; for a corner, the two exact one-sided slopes). The table is evidence that agrees with it, not a proof.

Moving $P$ changes $x$ and keeps $h$; moving $Q$ changes $h$. Arrow keys move the focused point by $0.25$, and the numeric fields accept two decimal places.

Derivation: why the secant slope approaches the derivative

For a polynomial the quotient simplifies before any limit is taken. With $f(x)=x^2$,

The cancellation of $h$ is legitimate only because $h\ne0$. The simplified expression $2x+h$ is continuous in $h$, so its limit as $h\to0$ is $2x$. In general, for a polynomial $p$, Taylor expansion gives

a polynomial in $h$ (the omitted terms carry $h^2$, $h^3$, …) whose constant term is $p'(x)$. The explorer prints this exact polynomial for the cubic: its constant term is the tangent slope and the remaining terms measure how far a finite secant is from the tangent.

Worked example

Take $f(x)=x^3/6-x$ at $x=1$. Then $f(1)=-5/6$, and for every $h\ne0$ the exact difference quotient is

With $h=1$ the secant slope is $1/6$; with $h=1/2$ it is $-5/24$; with $h=1/4$ it is $-35/96$. These approach $f'(1)=1^2/2-1=-1/2$. The replay below steps through the halving sequence and ends with the separately derived limit.

When no derivative exists

• Corner. For $|x|$ at $0$, $m(h)=|h|/h$ equals $1$ for $h>0$ and $-1$ for $h<0$. Both one-sided slopes exist, but they differ, so $f'(0)$ does not exist.
• Vertical tangent. For $\sqrt[3]{x}$ at $0$, $m(h)=h^{-2/3}\to+\infty$. The graph is smooth to the eye, but the slope is not a real number.
• Endpoint. $\sqrt x$ is defined only for $x\ge0$. At $0$ only $h>0$ is possible, and the right quotient $1/\sqrt h$ is unbounded.
• Jump. A function that is not continuous at $x$ cannot be differentiable there: from the broken side, $f(x+h)-f(x)$ does not shrink while $h$ does.

Every one of these is reported by the stage as an undefined derivative with its reason; none is replaced by a nearby number.

Executable reference code

The pedagogical reference uses exact Fraction arithmetic for the difference quotient of a polynomial and shows the one-sided slopes of $|x|$. The practical version tabulates secant slopes with NumPy and compares them with the analytic derivative.

"""Secant slopes as exact difference quotients (III.3). h must be nonzero; the derivative is a limit, never m(0)."""
from fractions import Fraction as F


def poly_value(coefficients, x):
    """coefficients[k] multiplies x**k (constant term first)."""
    return sum(c * x**k for k, c in enumerate(coefficients))


def secant_slope(f, x, h):
    if h == 0:
        raise ZeroDivisionError("h = 0: two equal points determine no secant")
    return (f(x + h) - f(x)) / h


cubic = [F(0), F(-1), F(0), F(1, 6)]           # x**3/6 - x
f = lambda x: poly_value(cubic, x)

# Exact slopes for h = 1, 1/2, 1/4, ...: they approach f'(1) = 1/2 - 1 = -1/2.
slopes = [secant_slope(f, F(1), F(1, 2**k)) for k in range(6)]
assert slopes[:3] == [F(1, 6), F(-5, 24), F(-35, 96)]
assert all(abs(b + F(1, 2)) < abs(a + F(1, 2)) for a, b in zip(slopes, slopes[1:]))

# |x| at 0: the one-sided slopes are exactly -1 and 1, so no derivative exists.
absolute = lambda x: abs(x)
right = {secant_slope(absolute, F(0), F(1, 2**k)) for k in range(8)}
left = {secant_slope(absolute, F(0), -F(1, 2**k)) for k in range(8)}
assert right == {1} and left == {-1}

try:
    secant_slope(f, F(1), F(0))
except ZeroDivisionError:
    pass
else:
    raise AssertionError("h = 0 must not produce a slope")

"""Tabulate secant slopes with NumPy and compare with the analytic derivative (III.3)."""
import numpy as np

f = lambda x: x**3 / 6 - x
fprime = lambda x: x**2 / 2 - 1          # derived analytically, not from the table

x = 1.0
h = 2.0 ** -np.arange(0, 12)              # nonzero steps only
slopes = (f(x + h) - f(x)) / h
gaps = np.abs(slopes - fprime(x))

# For this cubic m(h) = f'(x) + h/2 + h**2/6, so the gap shrinks roughly in proportion to h.
assert np.all(np.diff(gaps) < 0)
assert gaps[-1] < 1e-3
# Both sides agree for a smooth function.
left = (f(x - h) - f(x)) / (-h)
assert np.allclose(left[-4:], fprime(x), atol=3e-3)   # gap ≈ h/2 ≤ 2⁻⁹

Explore

Make $h$ negative and compare with positive $h$ of the same size. Move $P$ to a point where the cubic’s tangent is horizontal ($x=\sqrt2$ is not on the input grid; try $x=1.5$ and $x=1.25$ and watch the sign of the tangent slope). On Jump, compare the left and right columns of the table. On sin x, notice that every value is marked $\approx$.

Common mistakes

Substituting $h=0$ into the quotient produces $0/0$, which has no value. Cancelling $h$ is valid only while $h\ne0$. A derivative needs agreement from both sides; averaging two different one-sided slopes does not create one. A graph without a visible break can still fail to be differentiable (a vertical tangent), and a numerical table that looks like it settles is not a proof that a limit exists.

Numerical considerations

In exact rational arithmetic, smaller $h$ always brings the secant closer to the tangent of a polynomial. In binary64 it does not: $f(x+h)-f(x)$ subtracts two nearly equal numbers, and the rounding error is divided by a tiny $h$. Chapter III.14 measures this trade-off between truncation and rounding error. Here the halving table stops at $h/64$, far above that regime.

Engineering and ML connections

Every gradient used to train a model is a collection of derivatives. Automatic differentiation computes them from the operations of a program (III.10) rather than from secants. Finite-difference quotients remain the standard independent check of a gradient implementation, provided the step size and its error are understood.

Exercises

Summary and checkpoints

A secant slope is a quotient with $h\ne0$. The derivative is the limit of these slopes as $h\to0$ from both sides, when that limit exists and is finite. Corners, vertical tangents, endpoints and jumps are places where it does not. Next, III.4 reads the derivative as a rate of change and uses the tangent line as a local linear model.