Calculus

A function with a hole in it

$f(x)$

The output of our function when you feed in $x$.

$x^2$

"$x$ squared" — multiply $x$ by itself.

$x - 1$

Just the input minus one.

$\dfrac{\cdot}{\cdot}$

Division. The horizontal bar means divide the top by the bottom.

The problem. If you plug in $x = 1$, you get $\frac{0}{0}$, which is undefined — division by zero. The function has no output at $x = 1$. But for every $x$ close to 1 (like 0.99 or 1.01), the output is well-defined. What's it heading toward?

Canceling a common factor

$(x-1)(x+1)$

A product. This is "$x$ minus one" times "$x$ plus one". Expand it and you get $x^2 - 1$ — that's called the difference-of-squares identity.

$\dfrac{(x-1)(x+1)}{x-1}$

The top and bottom share a factor of $(x-1)$. When $x \ne 1$ this factor is non-zero, so we can cancel it.

$x + 1$

The simplified form — identical to the original everywhere except at $x = 1$, where the original is undefined and this one equals 2.

The intuition. The function is a perfectly smooth line $y = x + 1$ with a single missing pixel at $x = 1$. The limit "fills in" what that pixel ought to be. Limits are a formal way to look through the hole.

Average rate of change (a.k.a. secant slope)

$s(t)$

Your position as a function of time. Plug in a time, get back a location (say, meters from your front door).

$t$

The starting time.

$\Delta t$

A small time interval — "how much time has passed." The capital Greek delta means "change in."

$s(t + \Delta t) - s(t)$

How far you've moved during that interval (final position minus starting position).

$\dfrac{\cdot}{\Delta t}$

Distance divided by time — that's speed. Standard units would be meters per second.

Picture. Plot $s$ against $t$. Pick two points on the curve at times $t$ and $t + \Delta t$. Draw the straight line through them. This expression is the slope of that line — rise over run. It's called a "secant" line because it slices through the curve at two points.

The definition of the derivative

$s'(t)$

Read "s prime of t." A new function of $t$ whose value at each time is the instantaneous rate of change of $s$ at that time.

$\lim_{\Delta t \to 0}$

Take the limit as $\Delta t$ shrinks to zero. Remember: we never actually set $\Delta t$ equal to zero (that would be division by zero); we sneak up on it.

$\dfrac{s(t+\Delta t) - s(t)}{\Delta t}$

The average speed over the interval $[t, t+\Delta t]$ — the secant slope.

The geometric punchline. As $\Delta t$ shrinks, the two points you drew through get closer and closer. The secant line rotates and settles into a final position: the tangent line — the straight line that just grazes the curve at the single point $(t, s(t))$. The derivative is the slope of that tangent line. Instantaneous rate of change = slope of the tangent line. Two names for the same thing.

Definition of the derivative (generic form)

$f$

Any function of one real variable.

$f'(x)$

The derivative of $f$, evaluated at $x$. This is itself a function — plug in any $x$, get back a number.

$h$

The small change in input, same role as $\Delta x$ or $\Delta t$ above. Using a single letter saves space.

$f(x + h) - f(x)$

How much the output changed when we bumped the input by $h$.

$\dfrac{f(x+h) - f(x)}{h}$

Average rate of change over the interval from $x$ to $x+h$. Geometrically: slope of the secant line between those two points on the graph.

What the derivative "is." It's a machine. Feed in a function, get back a new function. The new function tells you the slope of the original at every point. If the original rises, the derivative is positive. If it falls, negative. If it has a peak or valley, the derivative is zero there. This last fact is why optimization uses derivatives — zero slope means "can't locally go up or down" which means "extremum."

Four ways to spell the same idea

$f'(x)$

Lagrange notation. A literal prime mark on the function name. Compact and standard.

$\dfrac{df}{dx}$

Leibniz notation. Looks like a fraction; isn't one, really, but behaves like one in many useful ways. The "$df$" and "$dx$" are meant to evoke infinitesimally small changes. Tells you what we're differentiating ($f$) with respect to what ($x$), which matters when more than one variable is around.

$\dot{f}(x)$

Newton's notation. The dot over the letter means "derivative with respect to time." Mostly used in physics.

$Df(x)$

Operator notation. $D$ is the "differentiation operator" — a function that takes a function and returns its derivative. Mostly used in advanced or computational settings.

Why the mess? Historical politics. Newton and Leibniz invented calculus in parallel, each with their own notation. British mathematicians stuck with Newton's dots for a century out of loyalty and fell behind the Continent as a result. Leibniz's $\frac{df}{dx}$ won on merit: it's more flexible, and it makes the chain rule look like a fraction cancellation. But everybody still uses $f'$ for brevity. So you will see all four, sometimes on the same page.

Power rule

$\dfrac{d}{dx}$

"The derivative with respect to $x$ of …" — an operator applied to what follows.

$x^n$

$x$ to the power $n$. For positive integer $n$, this is $x$ multiplied by itself $n$ times. For non-integer $n$, it's defined through exponentials.

$n \, x^{n-1}$

The result: the exponent drops out front as a coefficient and the new exponent is one less.

How to remember. "Bring the exponent down, subtract one from the exponent." So $\frac{d}{dx}(x^3) = 3x^2$, $\frac{d}{dx}(x^7) = 7x^6$, and $\frac{d}{dx}(x) = 1 \cdot x^0 = 1$. A constant $c$ can be written $c \cdot x^0$, and the rule gives $0 \cdot c \cdot x^{-1} = 0$ — the derivative of a constant is zero, which makes sense because a constant function has slope zero everywhere.

Linearity of the derivative

$f(x), g(x)$

Two functions of $x$.

$c$

A constant — a number that doesn't depend on $x$, like $3$ or $\pi$.

$f'(x) + g'(x)$

Differentiate each piece separately, then add.

Why it works. If two cars are moving, the rate of change of their combined position is the sum of their individual rates. Obvious when you phrase it that way. This linearity is why differentiation plays well with linear algebra — it's itself a linear operator.

Product rule

$f(x) \, g(x)$

The product of two functions. For example $x^2 \cdot \sin x$.

$f'(x) \, g(x)$

"First one's derivative times the second one, untouched."

$f(x) \, g'(x)$

"First one, untouched, times the second one's derivative."

Why this shape. Think of a rectangle of area $A = f \cdot g$, where both sides are changing. Bumping $f$ slightly grows the rectangle by a strip of width $g$ (area $\Delta f \cdot g$). Bumping $g$ grows it by a strip of width $f$ (area $f \cdot \Delta g$). There's also a tiny corner bit $\Delta f \cdot \Delta g$ — but in the limit, that corner is a second-order small quantity and vanishes. Only the two strips survive. That's the product rule.

Quotient rule

$\dfrac{f(x)}{g(x)}$

One function divided by another — assuming $g(x) \ne 0$, since division by zero is undefined.

$f'(x)\,g(x) - f(x)\,g'(x)$

Numerator: "derivative of top times bottom, minus top times derivative of bottom." Note the minus sign — the order matters here, unlike the product rule.

$g(x)^2$

Denominator: the bottom, squared.

Mnemonic. "Low dee-high minus high dee-low, over the square of what's below." You will say this sentence in your head for the rest of your life.

Chain rule (Lagrange form)

$f(g(x))$

A function of a function — first apply $g$ to $x$, then apply $f$ to the result.

$f'(g(x))$

The derivative of the outer function $f$, evaluated at the inner function's output $g(x)$. Read it as "$f$-prime, with the whole expression $g(x)$ plugged in where $x$ would normally go."

$g'(x)$

The derivative of the inner function at $x$.

$\cdot$

Multiplication. The chain rule is a product of two derivatives, not a sum.

Pictorial intuition. Imagine gears. The inner function $g$ turns the inside gear — a tiny bump in $x$ causes a bump of size $g'(x)$ in $u = g(x)$. That bump then drives the outer gear: a bump of $g'(x)$ in $u$ causes a bump of $f'(u) \cdot g'(x)$ in the final output. You multiply the gear ratios. That's the chain rule.

Chain rule (Leibniz form)

$u$

The intermediate variable — whatever the inner function returns. Here $u = g(x)$.

$\dfrac{df}{du}$

How fast $f$ changes per unit change in $u$ (the outer rate).

$\dfrac{du}{dx}$

How fast $u$ changes per unit change in $x$ (the inner rate).

$\dfrac{df}{dx}$

How fast $f$ changes per unit change in $x$, the quantity we want.

It's literally unit cancellation. Read the right side as "(change in $f$ per change in $u$) times (change in $u$ per change in $x$)." The "$du$"s visually cancel and you're left with "change in $f$ per change in $x$." Leibniz designed the notation exactly so that this would work.

Chain rule worked example

$\cos(x^2)$

The derivative of the outer function (sine) evaluated at the inner output. The derivative of sine is cosine.

$2x$

The derivative of the inner function ($x^2$), by the power rule.

$2x\cos(x^2)$

Their product. The answer.

Common bug. Beginners often stop at $\cos(x^2)$ and forget to multiply by the derivative of the inside. That missing factor is what separates someone who passes calculus from someone who doesn't. Always ask: "what's the derivative of what's inside?"

Linearization

$f(x_0)$

The value of the function at your chosen base point.

$f'(x_0)$

The slope of the tangent line there.

$(x - x_0)$

How far you've moved from the base point.

$\approx$

"Approximately equal to." The approximation is exact at $x_0$ and gets worse as you move away, but for small moves it's often excellent.

Why this matters. Every time you hear the phrase "first-order approximation" or "small perturbation," this is the formula being invoked. Physics builds entire simulations out of it. Numerical methods like Newton's method for root-finding (next) use it as their core step. When you see $f(x+h) \approx f(x) + h f'(x)$ in a paper, no new idea is being introduced — it's just linearization rearranged.

Newton's iteration

$x_n$

Your current guess for the root. $n$ is the iteration counter — $x_0$ is the initial guess, $x_1$ the first refinement, and so on.

$f(x_n)$

The function value at your current guess. You want this to be zero.

$f'(x_n)$

The slope of the tangent at your current guess.

$\dfrac{f(x_n)}{f'(x_n)}$

A correction term, derived by asking: "where does the tangent line cross zero?" Rise over run means the run you need is $-f(x_n)/f'(x_n)$.

Why it converges fast. When you're close to a root, Newton's method typically doubles the number of correct digits each iteration — quadratic convergence. It's the workhorse behind square-root and reciprocal instructions in hardware, and behind iterative solvers in numerical analysis. The catch: it needs a decent starting guess and a non-zero derivative nearby, or it can bounce off into the weeds.

Riemann sum

$\sum_{i=1}^n$

"Sum from $i = 1$ to $n$." The capital Greek sigma means "add up all these terms as the index $i$ runs from 1 to $n$." So $\sum_{i=1}^3 i = 1 + 2 + 3 = 6$.

$n$

The number of rectangles you're using. Bigger $n$ means thinner rectangles and a better approximation.

$\Delta x$

The width of each rectangle, equal to $(b-a)/n$.

$x_i^*$

The sample point inside the $i$-th subinterval. Typical choices: the left endpoint, the right endpoint, or the midpoint. For continuous $f$, it doesn't matter in the limit.

$f(x_i^*)$

The height of the $i$-th rectangle, which is the function's value at the sample point.

$f(x_i^*) \, \Delta x$

Height times width — the area of one rectangle.

$S_n$

The sum of those rectangle areas — an approximation to the true area under the curve.

Why this works. For modest $n$, you're missing some area (where the curve rises above a rectangle's top) and over-counting some (where the curve drops below). For huge $n$, those errors shrink faster than you can blink. The curve's wiggles get drowned by the sheer thinness of each rectangle.

The definite integral

$\displaystyle\int$

The integral sign — a stretched-out "S" for "sum." Leibniz picked this letter on purpose to remind you that integration is a kind of infinite sum.

$a, b$

The lower and upper limits of integration — the left and right endpoints of the interval over which you're adding up.

$f(x)$

The integrand — the function you're integrating.

$dx$

Think of this as "an infinitesimally thin slice of width $dx$." It's the limit version of $\Delta x$. The "$dx$" also tells you the variable of integration — in this case, $x$.

Picture in your head. $\int_a^b f(x)\, dx$ is the total area under the curve $y = f(x)$ between $x = a$ and $x = b$. If $f$ dips below the axis, the area there counts as negative — the integral is a signed area. If you integrate a velocity, you get a displacement. If you integrate a rate of flow, you get a total volume. Integration is the machinery for "adding up a rate over a duration to get a total."

Indefinite integral (antiderivative)

$F(x)$

Any function whose derivative is $f(x)$. For $f(x) = 2x$, one antiderivative is $F(x) = x^2$.

$C$

An arbitrary constant — the "constant of integration." Since the derivative of a constant is zero, $x^2 + 1$, $x^2 - 7$, and $x^2 + \pi$ are all antiderivatives of $2x$. We write the $+ C$ to indicate the whole family.

No $a, b$

The indefinite integral has no limits on the integral sign — that's how you tell it apart from a definite integral. It returns a function, not a number.

Easy to confuse, important to separate. Definite integral $\int_a^b f(x) dx$ is a number (an area). Indefinite integral $\int f(x) dx$ is a function (plus a constant). They are linked by the Fundamental Theorem of Calculus, which is the next section, and which explains why both things got named "integral." Until that theorem is on the table, the link is just asserted: "reversing differentiation" and "computing areas" happen to be the same operation. Nobody would've guessed.

FTC Part 1

$\int_a^x f(t)\, dt$

The area under $f$ from the fixed lower limit $a$ up to a variable upper limit $x$. As $x$ changes, so does the area; the result is a function of $x$.

$t$

A dummy variable of integration. It's just a placeholder for the input to $f$ as we add up. We use $t$ instead of $x$ here because $x$ is already being used as the upper limit — reusing the same letter would be confusing.

$\dfrac{d}{dx}$

Differentiate the whole expression with respect to $x$.

$f(x)$

The original integrand, evaluated at the upper limit.

Why it's true (sketch). Suppose you've already found the area from $a$ to $x$, and you nudge $x$ slightly to $x + h$. You add a new strip of area, approximately $f(x) \cdot h$ (height times thin width). Divide by $h$ to get the rate of change, and you get $f(x)$. Take the limit as $h \to 0$ and the approximation becomes exact. The rate at which the area is accumulating is whatever height the curve currently has.

FTC Part 2

$\int_a^b f(x)\, dx$

The definite integral — the exact, limit-based area under $f$ from $a$ to $b$.

$F$

Any antiderivative of $f$ — any function whose derivative is $f$. Several work; the $+C$ cancels when you subtract.

$F(b) - F(a)$

The difference of the antiderivative at the two endpoints. Sometimes written $F(x) \big|_a^b$ as shorthand.

Why this is a miracle. The left side is a horrifying infinite sum — slice into $n$ pieces, take a limit, etc. The right side is two function evaluations and a subtraction. The Fundamental Theorem says they're equal. Every time you compute an area in a homework problem by finding an antiderivative and subtracting, you're using it. Archimedes would have killed for this theorem.

u-substitution

$u$

A new variable that stands for the inner function — in this example, $u = x^2$.

$du$

The differential of $u$, equal to $\frac{du}{dx}\, dx$. It bookkeeps the scaling factor that changing variables introduces.

$\sin(u) + C$

The antiderivative of $\cos(u)$, since $\frac{d}{du}(\sin u) = \cos u$.

$\sin(x^2) + C$

Substitute $x^2$ back in for $u$ at the end — you want your answer in the original variable.

What it really is. The chain rule says $\frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$. u-substitution is that equation read right-to-left: if you see $\cos(x^2) \cdot 2x$ in an integral, you know the antiderivative is $\sin(x^2)$. Same rule, different direction.

Integration by parts

$u$ and $v$

Two functions of $x$. You choose which part of the integrand to call $u$ and which to call $dv$.

$dv$

The "differential of $v$" — whatever $v$ was, $dv = v'(x)\, dx$.

$du$

Similarly, $du = u'(x)\, dx$.

$uv - \int v\, du$

The boundary term $uv$ plus a new integral you hope is simpler than the one you started with.

The trade. You swap one integral ($\int u\, dv$) for another ($\int v\, du$). The trick is to pick $u$ and $dv$ so that the new integral is easier than the original. If you pick badly, you just go in circles. A common heuristic: let $u$ be whatever gets simpler when you differentiate (like $\ln x$ or $x^n$), and let $dv$ be whatever you can integrate easily (like $e^x\, dx$ or $\sin x\, dx$).

Integration by parts worked out

$x e^x$

The boundary term $uv$. Comes for free.

$\int e^x \, dx$

The new integral — much easier than the original. An antiderivative of $e^x$ is just $e^x$, since $e^x$ is the unique function equal to its own derivative.

$(x - 1)e^x + C$

The final answer, after cleanup. You can check it by differentiating: $\frac{d}{dx}[(x-1)e^x] = e^x + (x-1)e^x = x e^x$. Correct.

The pattern. Integration by parts is your tool for integrals where the integrand is a product of two different kinds of functions — polynomial times exponential, polynomial times trig, log times polynomial. It cuts the degree of one factor per application. Sometimes you have to apply it twice.

Taylor series

$f(a)$

The constant term — the value of $f$ at the expansion point $a$. Matches $f$ at $x = a$.

$f'(a)(x - a)$

The linear term. Adding this gives you the tangent line through $(a, f(a))$ — a first-order approximation.

$f''(a)$

The second derivative at $a$ — the derivative of the derivative. Measures how the slope itself is changing; geometrically, the concavity (curving up or down).

$2!$

"2 factorial," equal to $2 \cdot 1 = 2$. In general $n! = n(n-1)\cdots 1$.

$(x - a)^n$

The $n$-th power of "distance from the expansion point." For $x$ close to $a$, higher powers shrink fast, so higher-order terms contribute less and less.

$f^{(n)}(a)$

The $n$-th derivative of $f$ evaluated at $a$. Computed by differentiating $n$ times in a row.

$\cdots$

Continues forever. The full series has a term for every $n = 0, 1, 2, 3, \dots$

What's going on. You're building a polynomial that matches $f$ at $a$: same value, same slope, same curvature, same rate-of-change-of-curvature, and so on. Each new term fixes one more derivative. For many functions, if you include enough terms, the polynomial gets arbitrarily close to $f$ on some interval around $a$. Practical consequence: you can replace a complicated function with a few terms of its Taylor series and get a cheap, accurate approximation.

Maclaurin series (Taylor around 0)

$f(0), f'(0), f''(0), \dots$

The function and all its derivatives evaluated at $x = 0$.

$x, x^2, x^3, \dots$

The powers of $x$ itself, since $(x - 0)^n = x^n$.

Three series you'll meet everywhere. For $e^x$: $1 + x + x^2/2 + x^3/6 + \cdots$. For $\sin x$: $x - x^3/6 + x^5/120 - \cdots$. For $\cos x$: $1 - x^2/2 + x^4/24 - \cdots$. Each converges for all real $x$. These expansions are how calculators compute transcendental functions, and how physicists expand potentials near equilibrium, and how numerical analysts design integration schemes. When in doubt — Taylor expand.