The Derivative as the Slope of a Tangent Line

The derivative of a function at a point is, literally, the slope of the tangent line at that point. This visualization makes the abstract notion of $f'(x)$ feel tactile: as you slide a point along the parabola $f(x) = x^{2}$, the tangent line's slope updates in real time — and you can watch the derivative function $f'(x) = 2x$ trace itself out in the panel below.

The Physics — uh, the Math

The derivative is defined as the limit of the average slope as the interval shrinks to zero:

$f'(a) = \lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}$

For $f(x) = x^{2}$, plug in and simplify:

$f'(a) = \lim_{h \to 0}\dfrac{(a + h)^{2} - a^{2}}{h} = \lim_{h \to 0}\dfrac{2ah + h^{2}}{h} = \lim_{h \to 0}(2a + h) = 2a$

So $f'(x) = 2x$. At $x$, the slope of $y = x^{2}$ is $2x$.

Step-by-Step Solution

Given: $f(x) = x^{2}$. Find: the equation of the tangent line at the current point $x = a$, and verify it matches $f'(a) = 2a$.

---

Step 1 — Compute the derivative using the limit definition.

$f'(a) = \lim_{h \to 0}\dfrac{(a + h)^{2} - a^{2}}{h}$

Expand $(a + h)^{2} = a^{2} + 2ah + h^{2}$:

$f'(a) = \lim_{h \to 0}\dfrac{a^{2} + 2ah + h^{2} - a^{2}}{h} = \lim_{h \to 0}\dfrac{2ah + h^{2}}{h}$

Factor out $h$ in the numerator:

$f'(a) = \lim_{h \to 0}\dfrac{h(2a + h)}{h} = \lim_{h \to 0}(2a + h) = 2a$

Step 2 — Evaluate at the slider position. For example, at $a = 1$:

$f(1) = 1^{2} = 1, \qquad f'(1) = 2(1) = 2$

Step 3 — Write the tangent-line equation. Using point-slope form with point $(a, a^{2})$ and slope $2a$:

$y - a^{2} = 2a\,(x - a)$

$y = 2ax - 2a^{2} + a^{2} = 2ax - a^{2}$

For $a = 1$: $y = 2x - 1$. The tangent crosses the $x$-axis at $(\tfrac{1}{2}, 0)$ and passes through $(1, 1)$.

Step 4 — Verify with another point. At $a = 2$:

$f(2) = 4, \qquad f'(2) = 4, \qquad \text{tangent: } y = 4x - 4$

That tangent is steeper (slope 4) than at $x = 1$ (slope 2). The slope doubles as $a$ doubles — exactly as $f'(x) = 2x$ predicts.

---

Answer: The slope of the parabola $f(x) = x^{2}$ at $x = a$ is $f'(a) = 2a$, derived rigorously from the limit definition. As you slide the point right, the tangent line gets steeper at exactly twice the rate of $x$. The lower panel plots $f'(x) = 2x$ — a straight line through the origin with slope 2 — as the trace of all the slopes you've seen.

Try It

• Slide the x position widget — watch the cyan tangent line rotate as you move along the parabola.
• The slope readout in the HUD shows $2x$ live.
• Toggle show derivative graph to overlay $f'(x) = 2x$ in pink — the yellow dot tracks the current slope value.
• At $x = 0$, the tangent is horizontal (slope 0) — that's the bottom of the parabola.
• For negative $x$, the slope is negative — the tangent tilts the other way.