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Derivatives: from secants to a new function

First find a slope at one point. Then collect the slopes into a new function.

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01

How do we find the slope at a point on a curve?

Start with a familiar parabola, f(x) = x². Keep P fixed, choose Q on the same curve, and draw both lines.

−3−2−1123−2246810120xymPQ = 3.500mtangent = 2.000h = 1.5Δy = 5.25PQDrag P along the curveDrag Q; use the controls for precise movementf(x) = x²
P (x, f(x)) = (1, 1)Q (x+h, f(x+h)) = (2.5, 6.25)
Original functionSecant PQ · solidTangent at P · dashed
01

Read the two changes

h is the signed horizontal change from P to Q; it is negative when Q lies to the left.

Δx=h,Δy=f(x+h)−f(x)\Delta x=h,\qquad\Delta y=f(x+h)-f(x)
02

Use the two-point slope

Divide the vertical change by the horizontal change. This is still the secant slope.

mPQ=ΔyΔx=f(x+h)−f(x)hm_{PQ}=\frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{h}
03

Take the limit

If the slopes from both sides approach the same finite value, that value is f′(x).

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}h ≠ 0
Why does x² give exactly 2x?

Expand the square. Because h ≠ 0, cancel h first; then let h tend to zero. The finite animation suggests the result; this algebra establishes it.

(x+h)2−x2h=2xh+h2h=2x+h →h→0 2x\frac{(x+h)^2-x^2}{h}=\frac{2xh+h^2}{h}=2x+h\ \xrightarrow{h\to0}\ 2x
02

What if we record the slope at every x?

Left: the original function and its tangent. Right: an empty coordinate plane. Keep x unchanged and use the tangent slope as the new y-coordinate.

ORIGINAL FUNCTIONf(x) = x²
−2−11212345670xy(1, 1)
Tangent slope2
DERIVATIVE FUNCTIONf′(x) = ?
−2−112−5−4−3−2−1123450xyA blank plane. Each slope will become a height.
Same x. Use the slope as y.(x, f′(x))
1 / 5

At x = 1, the original function's instantaneous slope is 2. The derivative is that slope, so f′(1) = 2: record (1, 2) on the right.

x=1·x=2·x=0·x=-1·x=-2·

One derivative value. A whole derivative function.

f′(1) = 2 is a number: the slope at x = 1. f′(x) = 2x is a function: enter any x to get the slope there. The original height f(1) = 1 and the slope f′(1) = 2 are different quantities.

The final line is determined by f′(x) = 2x, derived above. A few points alone cannot uniquely determine a curve; the animation uses them to reveal the rule, not to prove it by sampling.