The Definition of the Derivative

The derivative \(f'(a)\) is the limit of the difference quotient \(\frac{f(a+h)-f(a)}{h}\) as \(h \to 0\) — one number that is at once the slope of the tangent line at \(a\) and the instantaneous rate of change of \(f\).

By the end you'll be able to compute \(f'(a)\) directly from the limit definition — expand, cancel the \(h\), take the limit — and read the result as both a tangent slope and an instantaneous rate of change.

Predict: at \(x = 1\) on \(y = x^2\), what slope does the secant line settle on as \(h\) shrinks to 0? Guess a number, then drag h toward 0 and watch the slope readout approach the tangent slope.

The secant passes through the fixed point \((a, f(a)) = (1, 1)\) and the moving point \((1+h, (1+h)^2)\). Its slope is exactly the difference quotient \(\frac{(1+h)^2 - 1}{h}\). As \(h \to 0\) (from either side — try negative \(h\) too), the secant pivots into the dashed tangent line. Hover or tab to either point for exact coordinates.

h = 1.50 · secant slope = [(1+h)² − 1]/h = 3.500 · tangent slope f′(1) = 2

y = x² — secant through (1, 1) and (1+h, (1+h)²), and the tangent it becomes
y = x² secant (slope = difference quotient) tangent at x = 1 (slope 2)

One limit, two readings: \(f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\) is the slope of the tangent line to the graph at \(a\), and it is the instantaneous rate of change of \(f\) at \(a\) — and you can compute it directly from this limit.

Intuitive

You can't draw a line through one point — a slope needs two. So borrow a second point a little distance \(h\) away and measure the slope of the secant line through \((a, f(a))\) and \((a+h, f(a+h))\). That slope is only an approximation, but it gets better as the borrowed point slides closer. In the limit, the secant pivots into the tangent line — exactly what you watched happen in the demo above as \(h\) shrank to 0.

Formal

The derivative of \(f\) at \(a\) is $$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, $$ provided this limit exists. An equivalent form pushes a moving point \(x\) toward \(a\): \(f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\). Note what plugging in \(h = 0\) would give: \(0/0\), a meaningless expression. The whole game is the one you already know from evaluating \(0/0\) limits — simplify algebraically while \(h \neq 0\) so the \(h\) cancels, and only then let \(h \to 0\).

Applied

The difference quotient \(\frac{f(a+h)-f(a)}{h}\) is an average rate of change over an interval of width \(h\) — like your average speed over a stretch of road. Shrink the interval and the average sharpens toward the instantaneous rate of change — the speedometer reading at one moment. Slope and rate of change are the same number seen from two angles: that's why one limit answers both "how steep is the graph here?" and "how fast is \(f\) changing right now?"

Worked example

Compute \(f'(a)\) for \(f(x) = x^2\) directly from the definition.
1. Form the difference quotient: \(\dfrac{(a+h)^2 - a^2}{h}\).
2. Expand the numerator: \(\dfrac{a^2 + 2ah + h^2 - a^2}{h} = \dfrac{2ah + h^2}{h}\).
3. Cancel \(h\) (valid because \(h \neq 0\) inside the limit): \(2a + h\).
4. Take the limit as \(h \to 0\): \(f'(a) = \) \(2a\).

Your turn

Same function, a specific point on the other side: find \(f'(-2)\) for \(f(x) = x^2\). Form the quotient: \(\dfrac{(-2+h)^2 - 4}{h} = \dfrac{4 - 4h + h^2 - 4}{h} = \dfrac{-4h + h^2}{h}\). Now cancel the \(h\) and take the limit as \(h \to 0\): \(f'(-2) = \) ____

Reveal the answer

Cancelling \(h\) leaves \(-4 + h\), and letting \(h \to 0\) gives \(f'(-2) = \) \(-4\) — matching the general formula \(2a = 2(-2) = -4\). A negative slope: on the left arm of the parabola, the tangent line points downhill.

More info — the same limit with a moving point x instead of a step h

The definition can be written without \(h\) at all: let a point \(x\) slide toward \(a\) and take \(f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\). It's the same secant-slope picture — the two points are now \((a, f(a))\) and \((x, f(x))\), and \(x - a\) plays the role of \(h\). Both forms are \(0/0\) at the target and both yield to the factor-and-cancel technique you learned for algebraic limits: for \(f(x) = x^2\), \(\frac{x^2 - a^2}{x - a} = \frac{(x-a)(x+a)}{x-a} = x + a \to 2a\). The OpenStax section in Dive deeper below derives both forms side by side.

Check your understanding

Question 1 of 4

What is the slope of the tangent line to \(y = x^2\) at \(x = 3\)?

Question 2 of 4

Why can't you find \(f'(a)\) by simply setting \(h = 0\) in the difference quotient \(\frac{f(a+h) - f(a)}{h}\)?

Question 3 of 4

The number \(f'(a)\) is simultaneously two things. Which pair?

Question 4 of 4

Use the limit definition to compute \(f'(2)\) for \(f(x) = x^2\), i.e. evaluate \(\lim_{h \to 0} \frac{(2+h)^2 - 4}{h}\).

Recap

  • \(f'(a) = \lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h}\) — the limit of the difference quotient, provided it exists. Equivalent form: \(\lim_{x \to a} \frac{f(x)-f(a)}{x-a}\).
  • The difference quotient is the slope of the secant through \((a, f(a))\) and \((a+h, f(a+h))\); as \(h \to 0\) the secant pivots into the tangent, so \(f'(a)\) is the slope of the tangent line at \(a\).
  • It is equally the instantaneous rate of change of \(f\) at \(a\): the limit of average rates over shrinking intervals.
  • Never set \(h = 0\) first — that's the meaningless \(0/0\). Simplify so the \(h\) cancels, then take the limit.
  • For \(f(x) = x^2\): \(f'(a) = 2a\); e.g. the tangent at \(x = 3\) has slope 6.

Dive deeper

Sources

  • The Derivative — Definition via the Difference Quotient