Calculus has two operations, both built on the limit. The derivative measures change at an instant, the slope of a tangent line. The integral adds up small pieces of change, the area under a curve. The Fundamental Theorem of Calculus shows that each undoes the other. Differential equations, infinite series and functions of several variables all build on those two operations. Most wrong answers come from structure rather than arithmetic: a missing chain-rule factor, a skipped endpoint, or a theorem used without its hypotheses.
Each chapter opens with the short version. Tap one to read the detail.
The function toolkit
~2 min
Read a domain from the formula as written, and split a composite by asking which operation you would do last. Logarithms need positive inputs, and calculus measures angles in radians.
Start with the domain. You drop any input that makes a denominator zero, puts a negative number under an even root, or gives a logarithm zero or a negative number. For f(x) = ln(x - 2)/sqrt(9 - x), the log needs x > 2 and the root needs x < 9, so the domain is the open interval from 2 to 9.
Composition is the second habit. In f(g(x)) the inner function g acts first, so the outer one is whatever you would compute last. In e^(3x), you multiply by 3 first and exponentiate last. A function has an inverse only when no two inputs share an output, which is why x^2 must be restricted to x of at least 0 before root x can undo it.
Logarithms turn products into sums and accept only positive inputs, so ln(x^2) equals 2 ln|x|, not 2 ln x. Angles are in radians because the clean derivative formulas for sine and cosine hold only in that unit.
The trap is simplifying before you read the domain. The expression x/x equals 1 everywhere except at 0, where it is undefined, and that missing point is exactly what a limit question will probe.
Rule: read the domain from the original formula, and name the outer function as the last operation you would perform.
Limits and continuity
~2 min
A limit describes outputs near a point, never the output at it, and 0/0 from substitution means more work is needed. Continuity means the value and the limit agree, and the big existence theorems need it on a closed interval.
A limit at a asks what the outputs approach as you move x toward a, from both sides. The value at a plays no part, and both one-sided limits must agree.
You can substitute directly wherever the function is continuous. When that produces 0/0, the limit can still exist. For (x^2 - x - 6)/(x - 3) at 3, factor the top as (x - 3)(x + 2). Cancel the shared factor, which is allowed because the limit never uses x = 3 itself, and substitute to get 5. When a bounded factor multiplies something that shrinks, you squeeze it: x^2 cos(1/x) stays between -x^2 and x^2, so it tends to 0.
A function is continuous at a when f(a) is defined, the limit exists, and the two are equal. To make a piecewise function continuous, you match the one-sided limits at the breakpoint. If f(x) = kx - 3 for x < 4 and root x for x of at least 4, then 4k - 3 = 2, so k = 5/4.
Continuity on a closed interval buys two guarantees. The Intermediate Value Theorem says every value between f(a) and f(b) is reached, so a sign change proves a root exists without locating it. The Extreme Value Theorem says a largest and a smallest value are both attained.
The trap is reading 0/0 as a final answer. It is a signal to simplify, not a verdict.
Rule: a limit is about nearby inputs, so cancel or bound the troublesome part and never use the value at the point itself.
What a derivative is, and the basic rules
~2 min
The derivative is the limit of average rates over shrinking intervals, so it is both a slope and a rate with units. Rewrite roots and fractions as powers first, and never treat the derivative of a product as the product of derivatives.
The derivative f'(a) is the limit of (f(a + h) - f(a))/h as h approaches 0. Each quotient is the slope of a chord, so in the limit you get the slope of the tangent. For f(x) = 3x^2 at 2, the quotient simplifies to 12 + 3h, which tends to 12. If you substitute h = 0 before simplifying, you always get 0/0.
A derivative is also a rate with units. If V(t) is liters in a tank and t is in minutes, V'(t) is in liters per minute at one instant. Differentiability implies continuity, but not the reverse: |x| is continuous at 0, yet its slopes are -1 and 1 on the two sides, so it has a corner and no derivative there.
The rules spare you the limit. The power rule gives n x^(n - 1) for any real exponent, so you write root x as x^(1/2) and 1/x as x^(-1) first. Rewriting often avoids a harder rule: (x^3 + 4)/x^2 becomes x + 4x^(-2), with derivative 1 - 8x^(-3). A product follows (fg)' = f'g + fg', and a quotient follows (f'g - fg')/g^2, in that order.
The trap is differentiating each factor and multiplying. For x^4 times x^5 that gives 20x^7, but the product is x^9, whose derivative is 9x^8.
Rule: linearity covers sums and constant multiples only; a product or quotient needs its own rule, or a rewrite that removes it.
The chain rule and what it powers
~2 min
Differentiate the outer function with the inner one left inside, then multiply by the inner derivative. The same rule drives implicit differentiation, related rates and the transcendental derivatives, and a tangent line gives quick estimates.
For y = F(g(x)), the derivative is F'(g(x)) times g'(x). For sin(5x^2), the outer function is sine and the inner is 5x^2, so the derivative is cos(5x^2) times 10x.
In radians, sin x gives cos x and cos x gives -sin x. The function e^x is its own derivative, b^x picks up a factor ln b, and ln(u) gives u'/u. In implicit differentiation, every term containing y picks up a factor dy/dx.
Related rates use time as the variable. You write the relation first, then differentiate with respect to t. For a balloon, V = (4/3) pi r^3 gives dV/dt = 4 pi r^2 dr/dt. If air enters at 50 cubic centimeters per second when r = 5 centimeters, dr/dt = 50/(100 pi), about 0.16 centimeters per second.
Near a point, you can let the tangent line L(x) = f(a) + f'(a)(x - a) stand in for the curve. Root 16.4 is about 4 + 0.4/8 = 4.05, against a true 4.0497. The estimate runs high because root x is concave down, bending below its tangent.
The trap is dropping the inner factor, or substituting the given instant's values before differentiating. Either way a changing quantity is treated as a constant, and a term vanishes.
Rule: name the layers first, keep one factor per layer, and put numbers in only after the last derivative is taken.
Extremes, the Mean Value Theorem and optimization
~2 min
Interior extremes occur only where f' is zero or undefined, but not every such point is one. On a closed interval, compare critical values with both endpoints. The Mean Value Theorem guarantees an instant whose rate equals the average rate.
A critical point is an interior point where f' is 0 or does not exist. Every interior local maximum or minimum occurs at one. The converse fails: x^3 has derivative 0 at 0 and keeps rising through it. For the absolute extremes of a continuous f on [a, b], you evaluate f at each critical point and at both endpoints, then compare.
The Mean Value Theorem says that if f is continuous on [a, b] and differentiable inside, some c in between has f'(c) = (f(b) - f(a))/(b - a). A car that covers 120 miles in 2 hours hits exactly 60 miles per hour at some moment. It promises that c exists, not where.
Optimization applies all of this. Suppose 160 meters of fence enclose a rectangle along a straight river, with no fence on the river side. If x is each side perpendicular to the river, the area is A = x(160 - 2x). Then A' = 160 - 4x is zero at x = 40, giving 40 by 80 meters, or 3,200 square meters. The domain runs from 0 to 80, and both endpoints give zero area, so the critical point is the maximum.
The trap is stopping at f'(c) = 0. A critical point can be a maximum, a minimum or neither until you justify it.
Rule: list critical points, add the endpoints of the meaningful domain, and compare values before naming an extreme.
Shape of a graph, asymptotes and L'Hopital's rule
~2 min
The sign of f' says where f rises and falls, and the sign of f'' says which way it bends. The bend switches only where f'' changes sign, and a zero of f'' just marks a place to look. L'Hopital's rule applies only to the forms 0/0 and infinity/infinity.
You mark every point where f' is zero or undefined, then test the sign of f' between those marks. A change from positive to negative marks a local maximum, and negative to positive a local minimum.
Take f(x) = 2x^3 - 3x^2 - 12x + 4. Here f'(x) = 6(x + 1)(x - 2), so f has a local maximum of 11 at x = -1 and a local minimum of -16 at x = 2. Next, f''(x) = 12x - 6 is negative below 1/2 and positive above, so the bend switches from concave down to concave up at x = 1/2.
The second derivative test is a shortcut at a critical point: f'' < 0 means a maximum and f'' > 0 a minimum. When f'' is 0 there, the test says nothing.
Far out, you divide top and bottom by the highest power in the denominator. Then (4x^2 + 1)/(7x^2 - x) tends to 4/7, a horizontal asymptote that the graph may still cross. For 0/0 or infinity/infinity, L'Hopital's rule replaces f/g by f'/g', differentiated separately: (e^x - 1)/(3x) at 0 becomes e^x/3, giving 1/3. Rewrite other forms, such as 0 times infinity, as a quotient first.
The trap is calling every zero of f'' an inflection point. The function x^4 has f''(0) = 0, yet it is concave up on both sides.
Rule: decide shape by sign changes across a point, and check the indeterminate form before every use of L'Hopital's rule.
Antiderivatives, the definite integral and the Fundamental Theorem
~2 min
An antiderivative is a whole family, F + C. A definite integral is a limit of Riemann sums that counts area below the axis as negative. The Fundamental Theorem links them: differentiating an accumulation returns the integrand, and F(b) - F(a) evaluates the integral.
You reverse the power rule: x^n integrates to x^(n + 1)/(n + 1) + C, except at n = -1, where 1/x gives ln|x| + C. The + C matters because antiderivatives on an interval differ only by a constant.
The definite integral from a to b is built from thin rectangles, added up as their widths shrink. Area below the x-axis counts as negative, so the result is a net amount.
The Fundamental Theorem of Calculus has two parts. Part one says that for continuous f, the integral of f from a fixed start to x has derivative f(x). With an upper limit of x^2, multiply by the chain factor: the integral of cos t from 0 to x^2 has derivative 2x cos(x^2). Part two says that if F' = f and f is continuous, the integral from a to b is F(b) - F(a). For 4x - 1 on [1, 3], 2x^2 - x gives 15 - 1 = 14, and the region, a trapezoid with heights 3 and 11 and width 2, confirms it.
When you integrate a rate, you get net change. Velocity gives displacement, while total distance needs the integral of speed, split where velocity changes sign.
The trap is evaluating across a point where the integrand blows up. For 1/x^2 on [-1, 2], -1/x gives -3/2, a negative answer for a positive integrand; the integral does not exist.
Rule: check continuity on the whole interval before subtracting, and add + C to every indefinite answer.
Substitution, parts and the other techniques
~2 min
Substitution undoes the chain rule, so look for an inner function whose derivative is present. Integration by parts undoes the product rule. Rational functions split into partial fractions, and trigonometric powers yield to identities.
Substitution works when an inner function's derivative also appears as a factor, perhaps off by a constant. In the integral of x^2 e^(x^3) dx, let u = x^3. Then du = 3x^2 dx, so x^2 dx is a third of du, and the answer is (1/3) e^(x^3) + C. Convert the limits too: on [0, 1], u also runs from 0 to 1, so the value is (e - 1)/3, about 0.573.
Integration by parts, the integral of u dv = uv minus the integral of v du, is the product rule read backward. You choose u as the factor that gets simpler when differentiated. For x cos x, u = x and dv = cos x dx give x sin x + cos x + C; the reverse choice raises the power of x.
The other tools match shapes. A rational function with distinct linear factors splits apart: 1/(x(x + 1)) = 1/x - 1/(x + 1), and each piece integrates to a logarithm. Powers of sine and cosine use sin^2 x + cos^2 x = 1 or the half-angle forms, and roots such as sqrt(a^2 - x^2) suggest a trigonometric substitution. Some integrands, such as e^(-x^2), have no elementary antiderivative at all.
The trap is mixing variables mid-problem. Plugging the old x-limits into an antiderivative written in u gives a wrong number with no warning.
Rule: try algebra, then substitution, then parts, then the shape-specific methods, and change the limits with the variable or not at all.
Area, volume and physical applications
~2 min
Slice the quantity into thin pieces you can measure, then integrate. Area between curves is top minus bottom, and a washer's area is pi(R^2 - r^2), never pi(R - r)^2. The same slicing gives work, mass and average value.
Every application starts by slicing: you decide what one thin piece contributes, then integrate across the range.
For area between curves, integrate top minus bottom, splitting wherever the curves swap places. Between y = 3x and y = x^2 on [0, 3], the line is on top, so the area is the integral of 3x - x^2, which is 9/2.
Revolve that region about the x-axis and every slice becomes a washer, with outer radius 3x and inner radius x^2. The volume is pi times the integral of 9x^2 - x^4, which is 162 pi/5. Squaring the difference instead gives 81 pi/10, a quarter of the true volume. When the axis is a line such as y = -2, you measure every radius from that line. For a region under y = f(x) turned about a vertical line, cylindrical shells avoid solving for x.
The same move handles physics. Work is force times distance only for a constant force. A spring's force kx grows with the stretch, so stretching it a distance s takes the integral of kx from 0 to s, which is (1/2)k s^2. Mass is the integral of a density, and the average value of f on [a, b] is its integral divided by b - a.
The trap is assuming one formula covers the whole region. When the top curve changes, or the curves cross, a single integral quietly cancels or miscounts.
Rule: sketch the region and one typical slice, write what that slice contributes, and only then set up the integral.
Differential equations
~2 min
A differential equation relates a function to its own derivatives. Check a solution by substituting it, find a family by integrating, and use an initial condition to pick one member. Separable equations, exponential change, cooling and logistic growth are the core models.
A solution is any function that makes the equation true on an interval; you check one by differentiating and substituting. Integrating leaves constants, so the general solution is a family of curves, and an initial condition selects one.
Take dy/dx = xy with y(0) = 2. The right side is a function of x times a function of y, so you separate: dy/y = x dx, then ln|y| = x^2/2 + C. That gives y = A e^(x^2/2), and the starting value fixes A = 2. Dividing by y discarded the constant solution y = 0, so check it separately; here it fails the initial condition.
A slope field draws short segments of the right slope, and solutions follow them. Euler's method steps along those slopes: from (0, 2) with step 0.1 the slope is 0, so it predicts y(0.1) = 2 against a true value of about 2.0100.
Three models recur. Growth proportional to size, dy/dt = ky, gives y = y0 e^(kt), with half-life ln 2/|k|; at k = -0.05 per year that is about 13.9 years. In Newton's law of cooling, the gap to the surrounding temperature decays exponentially. Logistic growth levels off at a carrying capacity K and grows fastest at K/2.
The trap is applying the initial condition before integrating, which leaves no constant for it to determine.
Rule: integrate first, then use the condition, and confirm by substituting the answer back into the equation.
Improper integrals, infinite series and convergence tests
~2 min
An infinite integral or sum means a limit, which may or may not be finite. Terms shrinking to 0 is necessary for a series to converge but never enough. Choose a test by the shape of the terms, and respect which way a comparison runs.
An integral out to infinity is the limit of integrals out to a finite b. The integral of 1/x^p from 1 to infinity converges exactly when p > 1. When the integrand blows up inside the interval, you split there and make each piece converge on its own. Numerical rules estimate the rest: trapezoids overestimate on a concave-up curve, and Simpson's rule needs an even number of subintervals.
A series is the limit of its partial sums. With first term a and ratio r, a geometric series sums to a/(1 - r) when |r| < 1, where a is the first term actually present. For 3(1/4)^n summed from n = 2, the first term is 3/16, so the sum is (3/16)/(3/4) = 1/4.
The harmonic series 1 + 1/2 + 1/3 + ... is the warning: its terms go to 0, yet its partial sums grow without bound. The p-series, the sum of 1/n^p, converges exactly when p > 1.
You pick tests in order. Terms that do not approach 0 mean divergence. Then spot geometric or p-series shapes, or compare with one. The ratio test suits factorials and powers, but a ratio limit of 1 decides nothing. An alternating series converges when its term sizes decrease to 0, and stopping early leaves an error no bigger than the first omitted term.
The trap is a comparison pointing the wrong way. Lying below a convergent series proves convergence, but lying below a divergent one proves nothing.
Rule: treat shrinking terms as a filter only, and confirm convergence with a test whose hypotheses you have checked.
Power series and Taylor series
~2 min
A power series converges on an interval around its center, and the endpoints need their own checks. A Taylor series matches every derivative at one point. Substituting into known series beats computing derivatives, and the remainder measures a truncation's error.
A power series in (x - c) converges inside a radius R and diverges outside it. The ratio test usually finds R but goes silent at the endpoints. For the sum of x^n/(n 3^n), the ratio tends to |x|/3, so R = 3. At x = 3 the series is the harmonic series and diverges; at x = -3 it is the alternating harmonic series and converges. The interval runs from -3, included, to 3, excluded.
The Taylor series of f about a has coefficients f^(k)(a)/k! on powers of (x - a), and a Maclaurin series is the case a = 0. A few cover most work: e^x is the sum of x^n/n!, cosine is 1 - x^2/2 + x^4/24 - and so on, 1/(1 - x) is the sum of x^n, and ln(1 + x) is x - x^2/2 + x^3/3 - and so on.
You substitute rather than differentiate. Replacing x with -x^2 in the series for e^x gives the series for e^(-x^2) at once, ready to integrate term by term.
When you truncate, you get an estimate with a measurable error. Cos 0.2 is about 1 - 0.02 = 0.98. The series alternates with shrinking terms, so the error is at most the next term, 0.2^4/24, about 0.00007. Series also settle limits: sin x - x begins with -x^3/6, so (sin x - x)/x^3 tends to -1/6.
The trap is reporting the radius as the answer. The two endpoints can behave differently.
Rule: find R with the ratio test, then test each endpoint on its own with a series test.
Beyond y = f(x): parametric, polar, vectors and several variables
~2 min
Parametric curves carry direction and timing, and their slope is dy/dt divided by dx/dt. Polar area is one-half the integral of r^2. In several variables, a partial derivative holds the other variables fixed, and the gradient points uphill fastest.
A parametric curve gives x and y as functions of t. Its slope is (dy/dt)/(dx/dt), so for x = t^2 + 1 and y = t^3 - 2t at t = 2 the slope is 10/4 = 5/2. The second derivative is not the ratio of second derivatives: you differentiate dy/dx with respect to t, then divide by dx/dt again. Polar coordinates use x = r cos(theta) and y = r sin(theta), and polar area is one-half the integral of r^2 d(theta), with limits that sweep the region exactly once.
The dot product multiplies matching components and adds them, and a dot product of zero means the vectors are perpendicular. The cross product of two vectors in space is perpendicular to both, and swapping the order flips its direction.
For f(x, y), the partial derivative with respect to x treats y as a constant. For f = x^2 y + 3y, the partials are 2xy and x^2 + 3, so at (1, 2) the gradient, the vector of partials, is (4, 4). The rate of change in a direction is the gradient dotted with a unit vector. Toward (3, 4), you first divide by its length 5, giving (12 + 16)/5 = 5.6. The gradient points in the steepest uphill direction and is perpendicular to level curves.
The trap is using a direction vector that is not unit length. Dotting with (3, 4) directly gives 28, five times too large.
Rule: normalize a direction before computing a directional derivative, and in every setting ask which variables are held fixed.
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