CLEP Calculus Quick Review

A limit describes the value a function approaches as the input approaches a point, whether or not the function is even defined there.

$_x→ a f(x) = L$ means $f(x)$ gets arbitrarily close to $L$ as $x$ approaches $a$; limit laws: $= f± g$, $= f· g$, $= f/ g$ (if $ g≠0$); a two-sided limit exists iff the left-hand and right-hand limits agree.

The value $f(a)$ (may not exist, or may differ) vs.\ the limit $_x→ af(x)$ (concerned only with behavior near $a$, not at $a$).

Assuming a limit fails to exist just because $f(a)$ is undefined, when the two-sided limit may still exist perfectly well (a removable discontinuity).

Limit = where you're headed, not where you land.

Most limits that produce the indeterminate form $0/0$ can be evaluated by simplifying algebraically first --- factoring, canceling, or rationalizing --- then substituting directly.

Factor-and-cancel for rational functions sharing a zero factor; multiply by the conjugate for radicals: $_x→0/sqrtx+4-2x=_x→0/xx(sqrtx+4+2)=/14$.

A $0/0$ indeterminate form (requires algebraic rewriting before substitution) vs.\ a limit that truly does not exist (jump discontinuity, unbounded oscillation).

Canceling a common factor from numerator and denominator and forgetting that the simplified function still has a hole at that $x$-value --- the limit value is unaffected, but the domain is not the same.

$0/0$ $arrow$ factor, cancel, or rationalize $arrow$ then plug in.

End behavior (limits at infinity) reveals horizontal asymptotes; limits that blow up near a finite $x$-value reveal vertical asymptotes.

For rational functions, compare degrees: degree(num) $<$ degree(denom) $→$ limit is $0$; equal degrees $→$ ratio of leading coefficients; degree(num) $>$ degree(denom) $→$ limit is $±∞$; a vertical asymptote occurs at $x=a$ where the denominator $→0$ but the numerator does not.

$_x→∞f(x)$ (horizontal asymptote, describes end behavior) vs.\ $_x→ af(x)=∞$ (vertical asymptote, function is unbounded near a finite point).

Dividing numerator and denominator by the wrong power of $x$, or forgetting the sign flip when $x→-∞$ inside a square root (since $sqrtx^2=-x$ for $x<0$).

Degree race decides the horizontal asymptote; zero denominator (not numerator) $→$ vertical asymptote.

A function is continuous at $a$ only when the function value, the limit, and their equality all hold together; losing any one piece produces a specific type of discontinuity.

$f$ is continuous at $a$ iff (1) $f(a)$ is defined, (2) $_x→ af(x)$ exists, and (3) $_x→ af(x)=f(a)$; removable discontinuity = a hole (limit exists but doesn't match $f(a)$); jump discontinuity = one-sided limits disagree; infinite discontinuity = vertical asymptote.

A removable discontinuity (fixable by redefining a single point) vs.\ a jump or infinite discontinuity (cannot be fixed by redefining one point).

Declaring a piecewise function continuous at a boundary just because each piece is individually a continuous formula, without actually checking that the one-sided limits match at the boundary.

Defined + limit exists + they match $arrow$ continuous.

The derivative measures instantaneous rate of change, defined as the limit of the average rate of change (a secant slope) as the interval width shrinks to zero.

$f'(x)=_h→0/f(x+h)-f(x)h$; equivalently $f'(a)=_x→ a/f(x)-f(a)x-a$; geometrically $f'(a)$ is the slope of the tangent line at $x=a$.

Average rate of change over $[a,b]$, $/f(b)-f(a)b-a$ (a secant slope) vs.\ instantaneous rate of change at a single point, $f'(a)$ (a tangent slope).

Forgetting to divide by $h$ before taking the limit, or failing to expand every instance of $x$ correctly inside $f(x+h)$.

Secant slope, shrink the interval $arrow$ tangent slope = derivative.

Differentiation is linear, so the derivative of a sum, difference, or constant multiple reduces to differentiating each piece separately.

$/ddx[x^n]=nx^n-1$ (Power Rule, any real $n$); $/ddx[cf(x)]=cf'(x)$; $/ddx[f(x)± g(x)]=f'(x)± g'(x)$; the derivative of a constant is $0$.

The Power Rule ($x^n$: exponent fixed, base is the variable) vs.\ exponential differentiation ($a^x$: variable is the exponent, base is fixed) --- these require different rules entirely.

Applying the Power Rule to $a^x$ (incorrectly producing $xa^x-1$) instead of the exponential rule $a^x a$, or forgetting to bring down a negative or fractional exponent.

Bring the exponent down, subtract one $arrow$ power rule; sums split apart.

The derivative of a product is not the product of the derivatives --- each factor's derivative must be paired with the other factor left undifferentiated.

$/ddx[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$.

Product Rule numerator ($f'g+fg'$) vs.\ the common wrong guess of simply multiplying the derivatives, $f'g'$.

Multiplying the two derivatives together instead of applying the full product rule, especially tempting when one factor is a simple polynomial that "looks easy."

First times derivative of second, plus second times derivative of first.

Differentiating a quotient requires a specific asymmetric formula in which both the order of subtraction and the squared denominator matter.

$/ddx≤ft[/f(x)g(x)]=/f'(x)g(x)-f(x)g'(x)[g(x)]^2$.

The correct numerator order ``low d-high minus high d-low,'' $f'g-fg'$, vs.\ the reversed (and wrong) order $fg'-f'g$, which flips the sign of the result.

Reversing the order of subtraction in the numerator, or forgetting to square the denominator.

``Low d-high minus high d-low, square the bottom and away we go.''

For a composite function, differentiate the outside function while keeping the inside intact, then multiply by the derivative of the inside.

$/ddx[f(g(x))]=f'(g(x))· g'(x)$; for nested compositions, chain together one derivative factor per layer.

The Chain Rule (for a composite function $f(g(x))$) vs.\ the Product Rule (for two separate factors multiplied together, $f(x)· g(x)$).

Forgetting to multiply by the derivative of the inner function --- ``forgetting the inside derivative'' --- especially on functions like $(3x)$ or $(2x+1)^5$.

Outside derivative $×$ inside derivative --- never skip the inside.

Each trig derivative pairs a ``co-'' function with a negative sign, and all six follow from the sine and cosine derivatives via the quotient rule.

$/ddx x= x$; $/ddx x=- x$; $/ddx x=^2x$; $/ddx x=-^2x$; $/ddx x= x x$; $/ddx x=- x x$.

$/ddx x=^2x$ vs.\ $/ddx x= x x$ --- easy to swap which formula produces which.

Dropping the negative sign on the derivatives of cosine, cotangent, and cosecant --- the three ``co-'' functions.

Co-functions get the minus sign: $- x$, $- x x$, $-^2x$.

$e^x$ is the unique function that is its own derivative; every other exponential or logarithmic derivative scales by a factor of $ a$.

$/ddxe^x=e^x$; $/ddxa^x=a^x a$; $/ddx x=/1x$; $/ddx_a x=/1x a$.

$/ddxe^x=e^x$ (no chain factor needed) vs.\ $/ddxe^g(x)=e^g(x)g'(x)$ (chain rule required whenever the exponent is not simply $x$).

Treating $e^x$ like a power-rule case (incorrectly writing $xe^x-1$) instead of recognizing it needs the exponential rule.

$e^x$ never changes; every other base picks up a factor of $ a$.

The derivatives of arcsine and arctangent are purely algebraic --- no trig functions appear --- which is exactly what makes them useful as antiderivative targets later.

$/ddx x=/1sqrt1-x^2$; $/ddx x=/11+x^2$; with the chain rule, $/ddx(g(x))=/g'(x)1+[g(x)]^2$.

$/ddx x=/1sqrt1-x^2$ (square root in the denominator, domain restricted to $(-1,1)$) vs.\ $/ddx x=/11+x^2$ (no square root, domain is all reals).

Forgetting the chain-rule factor $g'(x)$ when the argument is not simply $x$, e.g.\ differentiating $(3x)$ and leaving off the extra factor of $3$.

arcsin $→$ square root on the bottom; arctan $→$ plain sum of squares on the bottom.

When $y$ cannot be isolated, differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule to every $y$-term.

Differentiate term by term; every differentiated $y$-term picks up a factor of $/dydx$ (chain rule); then algebraically solve for $/dydx$. Example: $x^2+y^2=25 ⇒ 2x+2y/dydx=0 ⇒ /dydx=-/xy$.

Differentiating $y^2$ implicitly (gives $2y/dydx$, chain rule needed) vs.\ differentiating $x^2$ (gives $2x$, no chain factor since $x$ is the independent variable).

Forgetting to attach $/dydx$ to every differentiated $y$-term, or forgetting the product rule on mixed terms like $xy$ (which gives $y+x/dydx$).

Every $y$ gets a ``$+ dy/dx$'' tag; then isolate $dy/dx$.

Differentiating a derivative again produces the next-order derivative, each level carrying its own geometric or physical meaning along the chain position $→$ velocity $→$ acceleration.

Notation: $f''(x)=/d^2ydx^2$ is the derivative of $f'(x)$; for position $s(t)$, velocity $v(t)=s'(t)$ and acceleration $a(t)=v'(t)=s''(t)$.

$f''(x)$ (second derivative, rate of change of the slope; controls concavity) vs.\ $[f'(x)]^2$ (the square of the first derivative) --- similar-looking notation, entirely different quantities.

Stopping after one differentiation when a problem asks for the second derivative, or misapplying the product/quotient rule a second time without first simplifying the result.

Differentiate again $arrow$ one order higher; $s→ v→ a$.

The graphs of $f$, $f'$, and $f''$ are tightly linked --- the sign of $f'$ controls where $f$ increases or decreases, and the sign of $f''$ controls where $f$ curves up or down.

$f'(x)>0⇒ f$ increasing; $f'(x)<0⇒ f$ decreasing; $f'(x)=0$ at a critical point (candidate extremum); $f''(x)>0⇒ f$ concave up; $f''(x)<0⇒ f$ concave down.

A critical point of $f$ ($f'=0$ or undefined) vs.\ an inflection point of $f$ ($f''=0$ or undefined AND concavity actually changes) --- these are zeros of two different functions.

Being given the graph of $f'$ and answering ``where is $f$ increasing'' by tracking where the graph shown is rising, instead of where it is positive (above the $x$-axis).

$f'$ sign $→$ $f$'s direction; $f''$ sign $→$ $f$'s curvature.

Differentiability is a stronger condition than continuity --- every differentiable function is continuous, but continuous functions can still fail to be differentiable.

If $f$ is differentiable at $a$, then $f$ is continuous at $a$; the converse is false --- e.g., $f(x)=|x|$ is continuous but not differentiable at $x=0$ because its left- and right-hand derivatives disagree.

A corner or cusp (continuous, but the one-sided derivatives disagree) vs.\ a vertical tangent (continuous, but the derivative is unbounded) --- both fail differentiability for different graphical reasons.

Assuming a function is differentiable everywhere just because it ``looks continuous,'' without checking for corners, cusps, or vertical tangents.

Differentiable $⇒$ continuous, but not the reverse (think $|x|$ at $0$).

If a function is smooth enough on a closed interval, some interior point must have a tangent slope equal to the interval's overall average slope.

MVT: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $c(a,b)$ with $f'(c)=/f(b)-f(a)b-a$; Rolle's Theorem is the special case where $f(a)=f(b)$, guaranteeing some $c$ with $f'(c)=0$.

Rolle's Theorem (requires $f(a)=f(b)$, concludes $f'(c)=0$) vs.\ the Mean Value Theorem (no such requirement, concludes $f'(c)=$ the average slope, which need not be zero).

Applying MVT or Rolle's Theorem without verifying both hypotheses --- continuous on the closed interval AND differentiable on the open interval --- since a corner inside the interval breaks the guarantee.

Continuous + differentiable on interval $arrow$ some tangent matches (or, if endpoints equal, is flat).

When direct substitution into a limit gives an indeterminate form, differentiating the numerator and denominator separately --- not using the quotient rule --- can resolve it.

If $_x→ a/f(x)g(x)$ is $/00$ or $/∞∞$, then $_x→ a/f(x)g(x)=_x→ a/f'(x)g'(x)$, provided the right-hand limit exists; it may be reapplied if the new limit is still indeterminate.

L'H\^opital's Rule (differentiate numerator and denominator separately, $f'/g'$) vs.\ the Quotient Rule (a single derivative of the whole quotient, $/f'g-fg'g^2$) --- reaching for the Quotient Rule here is a common wrong instinct.

Applying L'H\^opital's Rule to a limit that is not actually in $/00$ or $/∞∞$ form, or forgetting to re-check the form before applying it a second time.

$0/0$ or $∞/∞$ only $→$ derivative of top over derivative of bottom, separately.

Near the point of tangency, a differentiable curve is well-approximated by its tangent line, giving a fast way to estimate function values nearby.

Tangent line at $x=a$: $y=f(a)+f'(a)(x-a)$; linear approximation $L(x)=f(a)+f'(a)(x-a)≈ f(x)$ for $x$ near $a$.

The tangent line equation (built from $f(a)$ AND $f'(a)$ at one specific point) vs.\ the derivative function $f'(x)$ alone (a formula valid at every $x$, not itself the equation of a line).

Plugging a value into $f'(x)$ and calling that the tangent line, instead of using it only as the slope and still needing the point $(a,f(a))$ to build the full line equation.

Tangent line = point-slope with slope $f'(a)$ at point $(a,f(a))$.

The sign of $f'$ on either side of a critical number classifies that critical number as a local max, local min, or neither.

Find critical numbers where $f'(x)=0$ or is undefined; $f'$ changes $+→-$ at $c ⇒$ local max at $c$; $f'$ changes $-→+$ at $c ⇒$ local min at $c$; no sign change $⇒$ neither.

A critical number (candidate for an extremum, found from $f'$) vs.\ an actual local extremum (confirmed only after testing the sign change of $f'$ around it).

Assuming every critical number is automatically a max or min, without testing the sign of $f'$ on both sides --- a critical number can be neither, e.g.\ $f(x)=x^3$ at $x=0$.

Critical number, test sign of $f'$ before/after $→$ classify as max, min, or neither.

The second derivative's sign gives an alternate, often faster, way to classify a critical point as a max or min by checking concavity directly at that point.

If $f'(c)=0$ and $f''(c)>0$, $f$ has a local minimum at $c$; if $f'(c)=0$ and $f''(c)<0$, $f$ has a local maximum at $c$; if $f''(c)=0$, the test is inconclusive (fall back to the First Derivative Test); an inflection point occurs where $f''$ changes sign.

The Second Derivative Test (classifies critical points via concavity, requires $f'(c)=0$ first) vs.\ locating inflection points (where $f''=0$ or undefined and concavity changes, unrelated to whether $f'=0$ there).

Applying the Second Derivative Test when $f''(c)=0$ and concluding ``neither,'' when the test is actually inconclusive and the First Derivative Test must be used instead.

$f''>0$ smiles (min); $f''<0$ frowns (max); $f''=0$ tells you nothing by itself.

Real-world max/min problems reduce to writing a single-variable objective function --- using a constraint to eliminate a variable --- then applying calculus to find its extreme value.

Steps: (1) write the quantity to optimize as a function, (2) use a given constraint to reduce to one variable, (3) find critical points via $f'=0$, (4) classify with the first/second derivative test, and (5) on a closed interval, also check the endpoints.

An unconstrained critical-point search (only interior critical points matter) vs.\ closed-interval optimization (the true max/min could also occur at an endpoint, so endpoints must be compared too).

Finding the critical $x$-value and reporting it as the final answer, when the question actually asked for the optimized quantity itself (e.g., the minimum cost, not the $x$ that produces it).

Write one-variable formula, differentiate, set to $0$, classify, check endpoints if closed.

When two or more quantities are linked by an equation and both change over time, differentiate the equation implicitly with respect to time to relate their rates.

Steps: (1) write an equation relating the variables, (2) differentiate both sides with respect to $t$ (chain rule attaches $/d(·)dt$ to every changing quantity), (3) substitute given numerical values only after differentiating, (4) solve for the requested rate.

Related rates (differentiate an equation with respect to time, every quantity a function of $t$) vs.\ plain implicit differentiation with respect to $x$ (differentiate with respect to $x$, treating $y$ as a function of $x$).

Substituting known numerical values into the equation before differentiating, which incorrectly turns a changing variable into a constant and drops its rate term entirely.

Relate, differentiate w.r.t.\ $t$, substitute numbers last, solve for the wanted rate.

Position, velocity, and acceleration form a derivative chain along a line, and the signs of velocity and acceleration together describe whether the object speeds up or slows down.

$v(t)=s'(t)$, $a(t)=v'(t)=s''(t)$; speed is $|v(t)|$; the object speeds up when $v$ and $a$ share the same sign, slows down when they have opposite signs; the object is momentarily at rest when $v(t)=0$.

Velocity (signed, indicates direction) vs.\ speed (always nonnegative, the magnitude of velocity) --- a ``decreasing velocity'' is not the same as ``slowing down'' when velocity is negative.

Concluding an object is slowing down just because acceleration is negative, without checking whether velocity is also negative (same sign as $a$ means speeding up, even with negative acceleration).

$v$ and $a$ same sign $→$ speeding up; opposite signs $→$ slowing down.

An antiderivative reverses differentiation; because the derivative of any constant is $0$, every antiderivative comes with a ``$+C$'' family of solutions.

$ x^n dx=/x^n+1n+1+C\ (n≠-1)$; $ /1x dx=|x|+C$; $ e^x dx=e^x+C$; $ x dx=- x+C$; $ x dx= x+C$; $ ^2x dx= x+C$.

The indefinite integral $ f(x) dx$ (a family of functions, includes $+C$) vs.\ the definite integral $_a^b f(x) dx$ (a single number, no $+C$ needed).

Forgetting the ``$+C$'' on an indefinite integral, or misapplying the power rule to $n=-1$ (which needs $|x|$, not the undefined $/x^00$).

Reverse the derivative rule, and never forget $+C$.

$u$-substitution reverses the Chain Rule --- spotting an inner function and its derivative (up to a constant) already inside the integrand lets you simplify to a basic integral.

Let $u=g(x)$, so $du=g'(x) dx$; then $ f(g(x))g'(x) dx= f(u) du$; for definite integrals, either convert the limits to $u$-values or substitute back to $x$ before evaluating.

Substitution (requires the derivative of the inner function, or a constant multiple of it, to already appear in the integrand) vs.\ an integrand with no matching $du$ present, where substitution is simply the wrong first move.

Forgetting to change the limits of integration to $u$-values after substituting in a definite integral (or forgetting to substitute back to $x$ before plugging in the original $x$-limits).

Spot $g(x)$ and its derivative together $→$ let $u=g(x)$, $du=g'(x) dx$.

A definite integral is defined as the limit of a Riemann sum --- the limiting value of the total signed area of thinner and thinner approximating rectangles.

$_a^b f(x) dx=_n→∞_i=1^n f(x_i^*) x$, where $ x=/b-an$; left, right, and midpoint sums use different sample points $x_i^*$ within each subinterval.

A left Riemann sum (uses each subinterval's left endpoint, tends to underestimate an increasing function) vs.\ a right Riemann sum (uses the right endpoint, tends to overestimate an increasing function).

Using the wrong $ x$ (dividing by the wrong number of subintervals), or selecting the wrong sample point --- left vs.\ right vs.\ midpoint --- for the sum requested.

Rectangles $→$ thinner and thinner $→$ limit = exact signed area.

Definite integrals obey linearity and interval-additivity rules that let you split, combine, or reverse the limits of integration algebraically.

$_a^b[f± g] dx=_a^b f dx±_a^b g dx$; $_a^b cf dx=c_a^b f dx$; $_a^b f dx=_a^c f dx+_c^b f dx$; $_a^b f dx=-_b^a f dx$; $_a^a f dx=0$.

A definite integral where $f(x)$ dips negative on part of the interval (signed area --- regions below the $x$-axis subtract) vs.\ total unsigned area between the curve and the $x$-axis (requires splitting at the zeros and taking absolute values first).

Computing $_a^b f(x) dx$ directly and reporting it as ``the area'' when $f$ dips below the $x$-axis, instead of splitting at the zeros and flipping the sign of the negative pieces.

Signed area from the integral; true area needs $|f(x)|$ or split-and-flip below the axis.

The Fundamental Theorem of Calculus bridges differentiation and integration --- Part 1 says differentiating an accumulation function returns the original integrand, and Part 2 gives the shortcut for evaluating definite integrals.

FTC Part 1: $/ddx_a^x f(t) dt=f(x)$ (chain-rule version: $/ddx_a^g(x) f(t) dt=f(g(x))· g'(x)$); FTC Part 2: $_a^b f(x) dx=F(b)-F(a)$, where $F'=f$.

FTC Part 1 (differentiating an integral, produces a function) vs.\ FTC Part 2 (evaluating a definite integral using an antiderivative, produces a number).

Forgetting the chain-rule factor $g'(x)$ when the upper limit of an FTC Part 1 problem is a function of $x$ rather than just $x$ itself, e.g.\ $/ddx_a^x^2 f(t) dt$.

Part 1: derivative of an integral gives back $f$; Part 2: $F(b)-F(a)$ evaluates it.

The average value of a continuous function over an interval is the height of the rectangle --- with that interval's width --- whose area equals the area under the curve.

$f_avg=/1b-a_a^b f(x) dx$; by the Mean Value Theorem for Integrals, a continuous $f$ actually attains this average value somewhere in $[a,b]$.

Average value of a function ($/1b-a_a^b f dx$, uses the integral) vs.\ the simple average of the two endpoint outputs $/f(a)+f(b)2$ (a different, generally unequal, quantity).

Forgetting to divide by $(b-a)$ after computing the definite integral, leaving just the total accumulated area instead of the average height.

Average value = (total area) $$ (interval width).

The area between two curves is the definite integral of (top curve minus bottom curve), integrated over the interval where that ordering holds.

$A=_a^b[top(x)-bottom(x)] dx$; if the curves cross within $[a,b]$, split the integral at each intersection point and swap which function is ``top'' in each piece.

Area between two curves (top minus bottom, a nonnegative result when set up correctly) vs.\ a single definite integral of one function (signed area, which can be negative).

Integrating (bottom $-$ top) by mistake on the part of the interval after the curves swap positions, producing a negative contribution that should have been the reversed subtraction.

Top minus bottom, split at every crossing point.

A differential equation relating a quantity to its own rate of change can often be solved by separating variables and integrating both sides; a rate proportional to the current amount gives exponential growth or decay.

Separable form: rewrite as $/1y dy=k dx$ and integrate both sides; the model $/dydt=ky$ has general solution $y=y_0e^kt$, where $y_0$ is the initial amount, $k>0$ growth, $k<0$ decay.

$/dydt=ky$ (rate proportional to the current amount, gives exponential $y_0e^kt$) vs.\ $/dydt=k$ (constant rate, gives linear $y_0+kt$) --- similar-looking setups with very different solution families.

Forgetting to use the initial condition to solve for the constant of integration (or for $y_0$), leaving an unresolved family of curves instead of the one specific solution the problem requested.

$y'=ky ⇒ y=y_0e^kt$ --- separate, integrate, apply the initial condition.

Formula sheet = the eight lines above, cold, before test day.

When two terms sound alike, ask which one needs a sign or a $+C$.

Existence theorems (IVT, EVT, MVT, Rolle's) guarantee a $c$ exists; the tests (1st/2nd derivative) classify what $c$ is.

Every derivative in this list reads backward as an antiderivative formula.

Picture $→$ one-variable function $→$ $f'=0$ $→$ classify/compare $→$ answer the question asked.

44 MCQ, $$90 min, split no-calculator/calculator; know your rules cold AND be able to reason about a graph.

Domain $→$ $f'$ chart (shape) $→$ $f''$ chart (curvature) $→$ end behavior $→$ sketch.

Simplify $→$ spot $u$ $→$ convert limits $→$ integrate $→$ $F(b)-F(a)$ $→$ sanity-check the sign.

When a rule ``sounds universal,'' check its hypotheses --- CLEP traps live in the fine print.