Derivative of y = x^2

Double the side of a square tile and its area doesn't just double — it quadruples. That runaway growth is x^2 in action: the area of a square climbs faster and faster the longer its side gets, and this page pins down exactly how fast at any moment.

So far every slope you've computed has been constant. A flat line gives 0, and the line y = x gives 1 — one number, valid everywhere, end of story. Today the story changes, because today we meet our first genuine curve:

f(x) = x^2

Stand anywhere on a parabola and look around. Near the bottom the ground is almost level — a gentle valley floor. Walk outward and it tilts, then rears up; far from the origin it's a cliff face. The steepness is different at every single point, so no single number can be "the slope of x^2". Whatever the derivative is, it must be something that changes as x changes — a formula, not a number.

And here is the astonishing part: that formula turns out to be almost absurdly simple. Infinitely many different slopes — one for each point of the parabola — are all captured at once by

\frac{d}{dx}\bigl[\,x^2\,\bigr] = 2x

Twice wherever you're standing. At x = 1 the slope is 2; at x = 10 it's 20; at x = -4 it's -8, running downhill. You've met this rule in real life, too: a dropped stone falls a distance proportional to t^2, which is exactly why falls get scarier so fast — after twice the time the stone isn't falling twice as fast by accident, it's the 2t in the derivative of t^2. This page earns that formula twice over: once with honest algebra, and once with a picture so good you'll never forget where the 2 comes from.

The proof: run the limit definition, honestly

No shortcuts — we feed f(x) = x^2 into the difference quotient and simplify with every step in the open.

Step 1 — expand f(x+h). Squaring the binomial (no skipped middle term!):

f(x + h) = (x+h)^2 = (x+h)(x+h) = x^2 + 2xh + h^2

Step 2 — subtract f(x). The x^2 terms cancel, and what remains is exactly the change in the function caused by the step h:

f(x+h) - f(x) = \bigl(x^2 + 2xh + h^2\bigr) - x^2 = 2xh + h^2

Step 3 — divide by h. Both surviving terms carry a factor of h, so it divides out cleanly. (This is legal precisely because h \neq 0 during the algebra — we shrink it later, we never set it to zero here.)

\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2}{h} = \frac{h\,(2x + h)}{h} = 2x + h

Step 4 — take the limit. The quotient is exactly 2x + h for every nonzero step: the true slope 2x, plus an error of exactly h. Shrink the step and the error shrinks with it — from either side:

f'(x) = \lim_{h \to 0}\,(2x + h) = 2x

Notice where each piece went. The 2xh term was proportional to the step, so dividing by h left the survivor 2x. The h^2 term was proportional to the step squared, so after dividing it still carried a factor of h — and the limit killed it. That split — the part proportional to h survives, everything of higher order dies — is the engine inside every derivative you will ever compute.

The picture that explains the 2

The algebra is airtight, but it doesn't show you why the answer is 2x rather than 3x or x. For that, remember what x^2 actually is: the area of a square with side x. Asking "how fast does x^2 grow?" is asking "how much area do I gain when the square's side grows a little?" — and a square has two edges that sweep out new area when its side grows. Step through it:

Grow the side from x to x + h and the new area arrives in three pieces: a strip of area x \cdot h down the right side, an identical strip of area x \cdot h across the top — there's your 2xh, one xh per edge — and the little h \times h corner where the strips would overlap, area h^2. That's the identity (x+h)^2 = x^2 + 2xh + h^2 drawn as furniture: big square, two strips, tiny corner.

Now divide the gained area by the step. The two strips give 2xh / h = 2x — a contribution that doesn't care how small the step was. The corner gives h^2 / h = h — a contribution that shrinks along with the step, and vanishes in the limit. The derivative 2x is literally the two edges of the square: a square of side 3 gains area through two edges of length 3, so it grows at rate 6. (File this picture away — a cube has three faces sweeping out volume, and sure enough x^3 will turn out to grow at rate 3x^2. The pattern is already whispering.)

Read the formula at specific points

A derivative formula is only useful if you can evaluate it, so let's take f'(x) = 2x for a walk along the parabola:

f'(3) = 2(3) = 6 \qquad f'(1) = 2 \qquad f'(0) = 0 \qquad f'(-2) = -4 \qquad f'(-10) = -20

Check each against the picture. At x = 3 the tangent rises six units for every one across — steep, and climbing. At x = 0 the tangent is perfectly flat: the bottom of the valley, the one point where the parabola pauses between falling and rising. And at x = -2 the slope is -4 — the minus sign isn't decoration, it says the curve is descending there: walk rightward on the left branch and you walk downhill. The sign of 2x encodes fall-or-rise, its size encodes how steeply, and both come free from one evaluation.

You can also run the formula in reverse: where is the slope equal to 10? Solve 2x = 10 to get x = 5. Where is it -6? At x = -3. Every slope you could ask for lives at exactly one point of the parabola — half the requested slope. Now watch all of this at once: slide the point and see the tangent tip over, its live slope always reading twice the x-value.

Around 1604 — some eighty years before Newton or Leibniz wrote a derivative — Galileo Galilei wanted to know how falling objects speed up. Free fall was too fast for the clocks of his day (he timed things with a water clock, weighing the water that dribbled out), so he slowed gravity down: he rolled bronze balls along a gently inclined plane, a groove lined with smooth parchment, and timed them over and over.

The pattern he found is beautiful. In equal ticks of time the ball covered distances in the ratio 1 : 3 : 5 : 7 : 9\ldots — the odd numbers. Add them up: after one tick the total distance is 1; after two, 1 + 3 = 4; after three, 1+3+5 = 9; after four, 16. Perfect squares! Total distance grows as t^2 — a law of nature, measured with dribbling water, that sat waiting decades for the mathematics that could explain it.

And Galileo's odd numbers are the derivative in disguise. The distance gained in the n-th tick is n^2 - (n-1)^2 = 2n - 1 — essentially 2n, twice the elapsed time: the speed of a falling object grows as 2t, exactly \tfrac{d}{dt}[t^2]. His experiment was begging for the 2x rule; calculus finally answered.

The curve and its derivative, side by side

There's one more way to see the rule, and it's the most important habit this page can leave you with: plot f and f' together. The solid curve is the parabola f(x) = x^2; the dashed line is its derivative f'(x) = 2x — a straight line through the origin with slope 2.

Read the two graphs against each other. Left of zero the parabola falls, and sure enough the dashed line sits below the axis — negative slopes. At x = 0 the parabola bottoms out flat, and the dashed line crosses through zero at that very instant. Right of zero the parabola climbs ever more steeply, and the dashed line rises to match, reporting a bigger slope at every step. The derivative is a complete running commentary on the original curve — where it says negative the curve descends, where it says zero the curve is momentarily level, where it grows the curve steepens. Learning to read that commentary is what the next pages are about.

See it explained