MISSION · Calculus
The Derivative as a Slope
How you find the slope of something that is never straight
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The idea in one sentence
A derivative is the slope of a curve at a single point, found by measuring slopes across shrinking gaps until the gap all but disappears.
The problem with curves
Slope needs two points. On a straight line that’s no trouble — every pair of points gives the same answer. On a curve, every pair gives a different one, and a single point gives none at all, because you cannot compute a rise over a run without a run.
So calculus takes the honest route. Pick your point at \(x\), pick a second point a small distance \(h\) further along, and find the slope of the line joining them. That line is called a secant, and its slope is a genuine average rate of change across the gap:
\[\frac{f(x + h) - f(x)}{h}\]Then shrink \(h\) and watch where that number heads.
Watching it converge
Take \(f(x) = x^2\) at \(x = 2\). Work the secant slope out for a few gaps:
| \(h\) | Secant slope |
|---|---|
| 1 | 5 |
| 0.5 | 4.5 |
| 0.1 | 4.1 |
| 0.01 | 4.01 |
The pattern is unmistakable: the slopes are marching towards 4. And the algebra says why. Expand the quotient for \(f(x) = x^2\):
\[\frac{(x + h)^2 - x^2}{h} = \frac{2xh + h^2}{h} = 2x + h\]Every secant slope is exactly \(2x + h\). Shrink \(h\) toward zero and what survives is \(2x\), which at \(x = 2\) is 4 — no approximation left in it.
\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 2x\]That limit is the derivative. The line it describes, touching the curve at one point and matching its direction, is the tangent.
What it tells you
The derivative is a function in its own right, and its sign reads like a report on the original:
- Positive — the curve is rising there.
- Negative — it’s falling.
- Zero — it’s momentarily flat, which is where peaks and valleys live.
Physically, if \(f\) is position over time, \(f'\) is velocity. If \(f\) is velocity, \(f'\) is acceleration. Any time one quantity changes with respect to another, the derivative is the rate of that change at an instant rather than across an interval.
Common mix-ups
- Setting \(h = 0\) instead of taking the limit. That gives \(\frac{0}{0}\), which is meaningless. Simplify the quotient first, then let \(h\) go to zero.
- Confusing average and instantaneous rates. The secant is the average across a gap; the tangent is the rate at a point.
- Reading a zero derivative as “the function is zero.” It means the slope is zero. \(f(x) = x^2 + 7\) has derivative zero at \(x = 0\), where \(f\) is 7.
- Expecting every curve to have a derivative everywhere. A sharp corner has no single tangent — approach from either side and you get different answers.
Check yourself
Using \(f'(x) = 2x\), what is the slope of \(y = x^2\) at \(x = -3\)? What does the sign tell you?
\(f'(-3) = -6\). The slope is negative, so the parabola is falling as you move right through that point — which fits, since \(x = -3\) is on the left branch, before the vertex.
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