Students often memorize $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ and still cannot say what the number is for. The derivative measures sensitivity. It says how fast the output changes when the input is nudged, at one moment, when that rate settles on a single finite value. The same number is the slope of the best local line to the graph at that point. A limit is what turns a run of average rates into that one slope. If the averages refuse to settle, the derivative does not exist there, even when a derivative formula looks tidy somewhere else on the graph.

This is a first college reading, and a bridge out of AP Calculus, not a substitute for the course. On the explainer library it is the first page of the calculus cluster. The wider map is calculus foundations. The limit language under the formula is epsilon delta intuition. When the curve is an equation rather than $y=f(x)$, the companion is implicit differentiation.

Average rate and instantaneous rate

The difference quotient $\frac{f(x+h)-f(x)}{h}$ is an average rate on the interval from $x$ to $x+h$. It divides the change in output by the change in input. Nothing about that quotient is instantaneous yet. Sending $h$ toward $0$ asks whether those averages settle. When they settle on one finite number, that number is $f'(x)$, the instantaneous rate.

Example Use a cart on a floor, as a model, not as a lab log. Let $t$ be seconds since the cart passed a mark, and let $s(t)=t^2$ be position in meters past that mark. The bare formula hides a coefficient. The $1$ in front of $t^2$ stands for $1$ meter per second squared, so the square of the time really does come out in meters. Keep that coefficient in mind when you read units. You do not see it written next to $t^2$.

From $t=3$ to $t=5$ the cart goes from $s(3)=9$ meters to $s(5)=25$ meters. The change in position is $16$ meters. The change in time is $2$ seconds. The average velocity is $16/2=8$ meters per second. That single number mixes a slower moment at the start of the interval with a faster moment at the end. The derivative of $t^2$ is $2t$, so the instantaneous velocity is $6$ meters per second at $t=3$ and $10$ meters per second at $t=5$. The average, $8$, sits between them. It is not a third fact about some new instant, except in one special way noted below.

Algebra makes the same point for every step $h\neq 0$:

$$\frac{s(3+h)-s(3)}{h}=\frac{(3+h)^2-9}{h}=\frac{6h+h^2}{h}=6+h.$$

The simplified quotient $6+h$ is the average velocity from time $3$ to time $3+h$, in meters per second. Put in $h=2$ and you recover the $8$ from the story. Put in a tiny $h$ and you are near $6$. The limit as $h$ approaches $0$ is $6$, and that is $s'(3)$. For a small positive $h$ the average is a little larger than $6$, which matches a cart that is still speeding up.

On this parabola, the average rate on an interval equals the instantaneous rate at the midpoint. From $3$ to $5$ the midpoint is $t=4$, and $s'(4)=8$. That shortcut belongs to quadratic models. It is not a license to replace every derivative by a long secant on some other function.

Linear approximation is the best local line

Differentiability at a point $a$ means there is a slope $m$ such that the error $f(a+h)-f(a)-mh$ shrinks faster than the step $h$ as $h$ approaches $0$. That $m$ is $f'(a)$. Near $a$, the graph tracks the line $y=f(a)+f'(a)(x-a)$ more closely than it tracks any other straight line through the point. The line is the linear approximation. People also call it the tangent line, which is the same object once the derivative exists.

At $t=3$ the cart is at $9$ meters with velocity $6$ meters per second. The best local line is

$$L(t)=9+6(t-3).$$

Read it as a prediction rule. Start at the known position, and add the instantaneous velocity times how far the clock moves. At $t=3.1$, one tenth of a second later, the line predicts $9+6(0.1)=9.6$ meters. The true position is $(3.1)^2=9.61$ meters. The error is $0.01$ meters. The step was $0.1$ seconds. The error is much smaller than the step.

The expansion $(3+h)^2=9+6h+h^2$ shows why. The line keeps $9+6h$ and drops $h^2$. The dropped piece is the whole error. Divide that error by the step and you get $h$ itself:

$$\frac{s(3+h)-s(3)-6h}{h}=h.$$

As $h$ approaches $0$, this ratio approaches $0$. That is what "faster than the step" means. A wrong slope does not pass the test. If you used $5$ instead of $6$, the error would be $h+h^2$, and the error divided by $h$ would approach $1$, not $0$. A nearby slope leaves a miss about the size of the step. Only $m=6$ makes the miss shrink faster than the step. The same local line is what Newton's method uses when it guesses a root, and what a short error bound uses when it replaces a curve by a straight line on a tiny interval.

What the derivative is not

A few misconceptions show up so often that they are worth retiring before the failure cases.

  • It is not the next value of the function. $s'(3)=6$ is a velocity in meters per second, not a position. One second later the cart is at $16$ meters, not at $6$.
  • It is not a percent change. From $9$ meters to $9.61$ meters, the change $0.61$ divided by the starting position $9$ is about $0.068$. That relative change is not a velocity. A relative rate would divide the derivative by the current value, when that value is not zero. At $t=3$, $6$ meters per second divided by $9$ meters is two thirds per second. That is a different measurement from $6$.
  • A zero derivative does not guarantee a maximum or a minimum. For $f(x)=x^3$, the derivative is $3x^2$, so $f'(0)=0$. Still, $f(0.1)=0.001$ and $f(-0.1)=-0.001$. The graph flattens for a moment and keeps rising. A horizontal tangent can be a flat step in a climb.
  • A derivative at a point is stronger than continuity there, and continuity of the derivative is stronger still. If $f'(a)$ exists, then $f$ is continuous at $a$. The corner below is continuous and still has no derivative, so continuity is not enough. A slope formula on each side of a point also does not finish the job. You still need the limit of the difference quotient at the point itself.
  • A difference quotient with a large step is an average, not "the derivative." The story with $h=2$ produced $8$ meters per second. Calling that number $s'(3)$ is the main blur this page is trying to undo.

When a derivative does not exist

Three pictures cover most first course failures. In each one, either $f(a)$ is the wrong kind of value for a tangent, or the difference quotients refuse to settle on one finite number. If the function is not even defined at the point, stop earlier. The quotient needs $f(a)$.

A corner

Let $f(x)=|x|$ and look at $x=0$. The function is continuous there: values near $0$ are near $0$, and $f(0)=0$. The slopes do not agree. For $h>0$,

$$\frac{|h|-0}{h}=1.$$

For $h<0$, the same quotient equals $-1$. The right hand averages settle on $1$. The left hand averages settle on $-1$. There is no single limit, so $f'(0)$ does not exist. A tangent line would have to fall with slope $-1$ on the left and rise with slope $1$ on the right. No one line does both. Continuity was not the missing piece. Agreement of the one sided rates was.

A jump

Define $f$ by two rules: $f(x)=0$ when $x<0$, and $f(x)=1$ when $x\ge 0$. At $0$ the outputs do not settle on $f(0)$. From the left they sit at $0$, while $f(0)=1$. That is a failure of continuity. Differentiability always implies continuity, so a derivative is already impossible here.

The quotients show the same wreck. For $h>0$, the difference quotient is $0$. For $h<0$, it equals $-1/h$, which grows without a finite limit as $h$ approaches $0$. There is no tangent line across a gap. The graph disconnects, and a finite slope has nothing consistent to follow.

A vertical tangent

Let $f(x)=x^{1/3}$, the cube root, and look at $x=0$. This function is continuous on the whole line, including at $0$, where $f(0)=0$. It does not turn a corner in the way absolute value does. The difference quotient is

$$\frac{h^{1/3}}{h}=\frac{1}{h^{2/3}}.$$

As $h$ approaches $0$ from either side, $h^{2/3}$ approaches $0$ through positive values, so the quotient grows without a finite limit. Secant lines through the origin and a nearby point become arbitrarily steep. The line you want would be vertical. A vertical line is not the graph of $y=f(0)+m(x-0)$ for any real number $m$. The derivative, which has to be that $m$, does not exist. Both sides tell the same story, and the story is "no finite slope," which is a different failure from the corner, where the two sides settled on two different finite slopes.

Pitfall. A central difference can hide a corner. For $|x|$ at $0$, the central recipe averages the left slope and the right slope and reports $0$ for any step it can evaluate. That $0$ is not a derivative. Forward difference reports $1$. Backward difference reports $-1$. When the one sided estimates disagree, do not promote the central number into $f'(0)$. A jump or a vertical tangent usually misbehaves in a louder way: the estimate grows as the step shrinks, instead of settling.

How to know you understood

  1. State $s'(3)$ with units, and say why the average from $t=3$ to $t=5$ is a different number.
  2. Build $L(3.1)$ by hand and compare it with $s(3.1)$. Name the error, and say why a slope of $5$ would not be the best local line.
  3. Give one corner, one jump, and one vertical tangent, in words, and say which of the three is still continuous.
  4. Say what a derivative is not, in one sentence you could tell a classmate who thinks $f'(a)$ is "the next $y$ value."

Practice on the numerical derivative calculator

The numerical derivative calculator estimates a difference quotient. It does not prove that a derivative exists, and it does not know units. If you are thinking of $x^2$ as the cart model, you attach "meters per second" yourself. The page also refuses a step larger than $1$, so the average over those two seconds stays on paper. Use the tool for a short step, where the secant has a chance of looking like the tangent.

Try it. Open the calculator on $x^2$ at $x_0=3$ with step $0.001$. Central difference should show $6$. For this parabola a central difference is exact for any step the formula can evaluate, because the $h^2$ errors on the two sides cancel. Forward difference should show about $6.001$, which is $6+h$. Change the step to $0.1$. Forward difference becomes $6.1$. That is the average rate from $3$ to $3.1$, not a broken tool. It matches the algebra $6+h$.

Then type abs(x) at $x_0=0$ with a small step. Forward difference gives $1$. Backward difference gives $-1$. Central difference gives $0$. Trust the disagreement, not the calm $0$. That is the corner from the section above, seen numerically.

After those two checks, come back to a function whose derivative you can compute by a rule, and only then believe an estimate on a function you cannot differentiate by hand. If the slope is hidden in an equation $F(x,y)=0$, this calculator is the wrong tool. Use implicit differentiation. If you want the epsilon and delta language under the limit in the definition, use epsilon delta intuition and, beside it, the numerical limit estimator as a picture rather than a proof.

Estimate a slope you already understand. Start with x squared at 3, where the true derivative is 6, and compare a forward difference with a central difference.

Open the numerical derivative calculator

A course shaped path that includes this page is the AP Calculus AB bridge. Later, if a rate depends on the current amount, the separable differential equations page is the next calculus cluster reading, not a substitute for this one. If Jacobian language feels empty, people are sent back here from the Jacobian explainer. The repair is the local line on this page.

Citations and further reading

  • Strang, G. Calculus. MIT OpenCourseWare, 18.01SC Single Variable Calculus. ocw.mit.edu
  • OpenStax, Calculus Volume 1, the chapters that define the derivative and that treat linear approximations. openstax.org
  • Paul’s Online Math Notes, the definition of the derivative. tutorial.math.lamar.edu
  • Paul’s Online Math Notes, linear approximations. tutorial.math.lamar.edu
  • 3Blue1Brown, “The paradox of the derivative,” a video picture of an instantaneous rate. YouTube

Last reviewed: 24 September 2026.