What a derivative measures when you zoom in

What does the number f′(a) tell you about the graph of f?

Zoom in on the graph around the point (a, f(a)), with the same scale on both axes. When the graph straightens into a line that is not vertical, the slope of that line is f′(a). Stage 1 asks you which points straighten; the rest of the page turns the picture into a limit, an approximation and a proof.

Before you start · 3 quick checks on slopes, function values and |x|, about 2 min

Definitions

Let \(f\) be defined on an open interval that contains \(a\). For \(h \ne 0\), the difference quotient

\[ \frac{f(a+h) - f(a)}{h} \]

is the slope of the chord from \((a, f(a))\) to \((a + h, f(a + h))\). The function \(f\) is differentiable at \(a\) when the limit

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]

exists and is a real number. The one-sided derivatives use only \(h \gt 0\):

\[ f'_+(a) = \lim_{h \to 0^+} \frac{f(a+h) - f(a)}{h}, \] \[ f'_-(a) = \lim_{h \to 0^+} \frac{f(a) - f(a-h)}{h}, \]

when these limits exist. \(f'(a)\) exists exactly when both one-sided derivatives exist, are real numbers, and are equal.

Why the zoom keeps the same scale on both axes

The zoom in stage 1 shows the window \(a - h \le x \le a + h\), and the vertical range is chosen so that one unit up is as long on the screen as one unit across. Then a slope of 1 looks like 45°, and every chord slope you see is the true one. Stretching one axis more than the other would change every slope in the picture.

The left chord goes from \((a - h, f(a - h))\) to \((a, f(a))\), with slope \(\dfrac{f(a) - f(a-h)}{h}\); the right chord goes to \((a + h, f(a + h))\), with slope \(\dfrac{f(a+h) - f(a)}{h}\). Zooming in by a factor of 10 divides \(h\) by 10.

1 See

Zoom in on a point you choose

Choose a function and move the point along its graph: drag it, click on the graph, or focus the point and use the arrow keys. Then say what the graph will look like when the page zooms in on that point. After you answer, each press of Zoom in divides the window's half-width \(h\) by 10, and the table lists the slopes of the two chords from the point to the edges of the window.

Function

Chord slopes at each zoom, rounded to 4 decimals. Both chords are taken in the direction of increasing x, so a straight line gives a change in direction of 0° (the angle at the point between the two pieces would be 180° minus it); rounded to 0.01°.
Half-width \(h\)	Left chord slope \(\frac{f(a) - f(a-h)}{h}\)	Right chord slope \(\frac{f(a+h) - f(a)}{h}\)	Change in direction between the chords

Four zoom levels only suggest what happens. Whether \(f'(a)\) exists is decided by the limit as \(h \to 0\), as in the definition on the Start tab.

2 Try

The best straight-line approximation

Near a point where \(f\) has a derivative, the tangent line \(L(x) = f(a) + f'(a)(x - a)\) gives the linear approximation \(f(a + h) \approx f(a) + f'(a)h\). Its error is \(E(h) = f(a + h) - f(a) - f'(a)h\). How fast does that error shrink?

Error of \(1 + h\) as an approximation of \(e^h\), to 3 significant digits; the last column to 4 decimals.
\(h\)	\(E(h) = e^h - 1 - h\)	\(E(h)/h\)	\(E(h)/h^2\)

Goals: find the slope that hugs the curve

The line through \((a, f(a))\) with slope \(m\) is \(y = f(a) + m(x - a)\). Its error at \(x = a + h\) is \(E(h) = f(a + h) - f(a) - mh\). Move \(m\); the table shows \(E(h)\) and \(E(h)/h\) for six values of \(h\).

Function and point

Errors of the line with slope m: \(E(h)\) to 3 significant digits, \(E(h)/h\) to 4 decimals.
\(h\)	\(E(h)\)	\(E(h)/h\)

Continuity and derivatives

Key point The tangent line is the one line whose error, divided by \(h\), tends to 0. With any other slope \(m\) the ratio \(E(h)/h\) tends to \(f'(a) - m\). At a corner no single slope works from both sides.

3 Prove

A derivative forces continuity

Warm-up: the slope of \(x^2\)

Claim. For \(f(x) = x^2\), \(f'(a) = 2a\) for every real \(a\). Build the computation from the difference quotient \(\dfrac{(a+h)^2 - a^2}{h}\), one step at a time.

Theorem (differentiable implies continuous)

Let \(f\) be a function defined for every real number, and let \(a\) be a real number. For \(h \ne 0\) write

\[ q(h) = \frac{f(a+h) - f(a)}{h}. \]

Assume \(f\) is differentiable at \(a\): \(q(h)\) tends to a real number \(f'(a)\) as \(h \to 0\).

Conclusion. \(f\) is continuous at \(a\): \(f(a + h) \to f(a)\) as \(h \to 0\).

The proof uses the limit laws for sums and products: if \(u(h) \to U\) and \(v(h) \to V\) with \(U\) and \(V\) real, then \(u(h) + v(h) \to U + V\) and \(u(h)\,v(h) \to UV\).

The converse is false

A function can be continuous at \(a\) without a derivative there. \(f(x) = \lvert x\rvert\) is continuous at 0, but its quotient \(\dfrac{\lvert h\rvert}{h}\) is 1 for \(h \gt 0\) and \(-1\) for \(h \lt 0\): the one-sided derivatives are 1 and \(-1\). \(f(x) = \sqrt[3]{x}\) is continuous at 0, but its quotient \(\dfrac{\sqrt[3]{h}}{h} = h^{-2/3}\) grows without bound.

Key point For \(h \ne 0\), \(f(a+h) - f(a) = q(h)\cdot h\). A derivative makes \(q(h)\) tend to a real number, so the product tends to \(f'(a)\cdot 0 = 0\), which is continuity. Continuity alone does not guarantee a finite limit for \(q(h)\).

4 Apply

Problems

Each problem has hints and a full solution.