AP Calculus AB Quick Review

A limit describes the value a function approaches as the input approaches a point, regardless of the function's actual value there.

$_x→ c f(x) = L$ means $f(x)$ gets arbitrarily close to $L$ as $x$ gets close to $c$ from both sides. The limit exists only if $_x→ c^- f(x) = _x→ c^+ f(x) = L$. Use tables of values approaching $c$ from each side to estimate numerically.

$_x→ c f(x)$ (the limit, describing behavior near $c$) vs.\ $f(c)$ (the actual function value at $c$) --- these can differ or one can fail to exist while the other doesn't.

Students read a removable discontinuity (open circle) on a graph and report the $y$-value of the open circle as $f(c)$, or they assume the limit equals $f(c)$ without checking the graph for a hole or jump.

Limit = where it's headed $arrow$ not necessarily where it lands.

Most limits of continuous combinations of functions can be evaluated by direct substitution; when that gives $/00$, algebraic manipulation reveals the true limit.

$_x→ c[f(x)± g(x)] = _x→ cf(x) ± _x→ cg(x)$; $_x→ c[f(x)g(x)] = _x→ cf(x)·_x→ cg(x)$; $_x→ c/f(x)g(x) = /_x→ cf(x)_x→ cg(x)$ if $_x→ cg(x)≠ 0$. For $/00$ forms, factor and cancel, or rationalize (multiply by the conjugate).

A $/00$ indeterminate form (further algebra needed, limit may exist) vs.\ a nonzero-over-zero form like $/50$ (limit is $±∞$ or does not exist, no cancellation possible).

Students plug in $c$, get $/00$, and incorrectly conclude the limit does not exist instead of factoring/rationalizing to simplify first.

$/00$ $arrow$ factor or rationalize, don't quit.

Infinite limits describe unbounded behavior near a vertical asymptote; limits at infinity describe end behavior and horizontal asymptotes.

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

A limit that equals $∞$ (describes unbounded growth, technically ``does not exist'' but is reported as $∞$) vs.\ a limit that simply ``does not exist'' with no infinite trend (e.g., oscillation or a jump discontinuity).

Students forget to check the sign of the denominator on each side of a vertical asymptote, so they miss that a one-sided limit is $+∞$ while the other is $-∞$, making the two-sided limit DNE.

Degree race: bottom wins $→ 0$, tie $→$ ratio of leading coeffs, top wins $→ ±∞$.

Continuity requires the limit, the function value, and their equality to all line up at every point; the Intermediate Value Theorem exploits continuity to guarantee solutions exist.

$f$ is continuous at $x=c$ iff (1) $f(c)$ is defined, (2) $_x→ cf(x)$ exists, and (3) $_x→ cf(x) = f(c)$. IVT: if $f$ is continuous on $[a,b]$ and $k$ is between $f(a)$ and $f(b)$, then there exists $c$ with $f(c)=k$.

Removable discontinuity (limit exists but $≠ f(c)$, or $f(c)$ undefined --- a ``hole'') vs.\ jump/infinite discontinuity (limit does not exist at all --- cannot be fixed by redefining one point).

On IVT free-response questions, students state that a solution ``exists'' without first justifying that the function is continuous on the closed interval, losing the justification point.

IVT = continuous + value trapped between endpoints $arrow$ guaranteed hit.

The derivative is the instantaneous rate of change, defined as the limit of average rates of change (slopes of secant lines) as the interval shrinks to zero.

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

Average rate of change $/f(b)-f(a)b-a$ (slope of secant line over an interval) vs.\ instantaneous rate of change $f'(c)$ (slope of tangent line at one point).

Students asked to evaluate a limit of a difference quotient fail to recognize it as $f'(c)$ in disguise and try to compute it directly instead of identifying the underlying function and point.

Secant shrinks to tangent $arrow$ derivative is the limit of slopes.

Differentiability is a stronger condition than continuity: every differentiable function is continuous, but not every continuous function is differentiable.

$f$ is differentiable at $c$ only if $f$ is continuous at $c$ AND the left-hand and right-hand derivatives are equal (no corner, cusp, vertical tangent, or discontinuity at $c$).

Continuous but not differentiable (e.g., $f(x)=|x|$ at $x=0$, a corner) vs.\ differentiable implies continuous (the one-way implication that always holds).

Students assume continuity guarantees differentiability, missing that a sharp corner or vertical tangent line breaks differentiability even where the function is perfectly continuous.

Differentiable $⇒$ continuous, but NOT continuous $⇒$ differentiable.

Basic differentiation rules let you compute derivatives of polynomials and sums of power functions term by term without returning to the limit definition.

$/ddx[x^n] = nx^n-1$; $/ddx[c] = 0$; $/ddx[f(x)± g(x)] = f'(x)± g'(x)$; $/ddx[c· f(x)] = c· f'(x)$.

The power rule (exponent comes down, exponent decreases by 1: $nx^n-1$) vs.\ exponential differentiation (base stays, involves $$: $/ddx[a^x]=a^x a$) --- students often apply the power rule to $2^x$ by mistake.

Students forget to rewrite radicals and reciprocals as rational exponents (e.g., $sqrtx=x^1/2$, $/1x^3=x^-3$) before applying the power rule, leading to sign or exponent errors.

Bring the power down, subtract one from it.

The transcendental functions have fixed derivative formulas that must be memorized cold, since they cannot be derived quickly on exam time.

$/ddx[ x]= x$, $/ddx[ x]=- x$, $/ddx[ x]=^2 x$, $/ddx[ x]=-^2 x$, $/ddx[ x]= x x$, $/ddx[ x]=- x x$; $/ddx[e^x]=e^x$, $/ddx[a^x]=a^x a$, $/ddx[ x]=/1x$, $/ddx[_a x]=/1x a$.

$/ddx[ x]= x$ (no negative) vs.\ $/ddx[ x]=- x$ (negative sign) --- the sign only appears on cosine's derivative, a constant source of slips.

Students drop the $ a$ factor when differentiating general exponentials $a^x$ (treating them like $e^x$), or drop the negative sign on $/ddx[ x]$ and $/ddx[ x]/[ x]$.

Sine to cosine, cosine to negative sine --- co-functions carry the minus sign.

Near the point of tangency, a differentiable function can be closely approximated by its tangent line --- the basis of local linear approximation.

Tangent line at $x=a$: $y = f(a) + f'(a)(x-a)$. Local linearization approximates $f(x)≈ f(a)+f'(a)(x-a)$ for $x$ near $a$.

Tangent line approximation (uses the derivative at ONE point to estimate nearby values, generally most accurate very close to that point) vs.\ the actual function value (tangent line approximations can overestimate or underestimate depending on concavity).

Students use the tangent line approximation without checking concavity, so they don't realize/report whether the approximation over- or under-estimates the true value (concave up $→$ tangent line underestimates; concave down $→$ overestimates).

Tangent line: point-slope with $f(a)$ and $f'(a)$ --- zoom in enough and the curve looks straight.

To differentiate a composite function, differentiate the outer function (leaving the inner alone) and multiply by the derivative of the inner function.

If $y=f(g(x))$, then $y' = f'(g(x))· g'(x)$. Equivalently in Leibniz notation, $/dydx=/dydu·/dudx$. Chain rule can nest multiple times for compositions of three or more functions.

Product rule (two factors multiplied together, e.g., $x^2 x$) vs.\ chain rule (one function plugged inside another, e.g., $(x^2)$) --- students must first identify whether the expression is a product or a composition.

Students differentiate the outer function correctly but forget to multiply by the derivative of the inner function entirely, e.g., writing $/ddx[(3x)] = (3x)$ instead of $3(3x)$.

Outside derivative $×$ inside derivative --- never forget the inside's derivative.

Derivatives of products and quotients require dedicated rules because the derivative of a product/quotient is NOT the product/quotient of the derivatives.

Product rule: $(fg)' = f'g + fg'$. Quotient rule: $≤ft(/fg)' = /f'g - fg'g^2$.

Product rule $(fg)'=f'g+fg'$ (symmetric, order doesn't matter, no subtraction) vs.\ quotient rule $≤ft(/fg)'=/f'g-fg'g^2$ (order matters in the numerator's subtraction, and denominator is squared).

Students reverse the order of terms in the quotient rule numerator (writing $fg'-f'g$ instead of $f'g-fg'$), or forget to square the denominator.

Quotient rule: ``low d-high minus high d-low, over the square of what's below.''

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

Differentiate each term with respect to $x$; every term containing $y$ picks up a factor of $/dydx$ via the chain rule (e.g., $/ddx[y^2] = 2y/dydx$). Then solve algebraically for $/dydx$.

Implicit differentiation (differentiating an equation relating $x$ and $y$ where $y$ is not isolated, requires $/dydx$ on $y$-terms) vs.\ explicit differentiation (function already solved for $y=f(x)$, no extra $/dydx$ factor needed).

Students forget to attach $/dydx$ when differentiating $y$-terms (e.g., differentiating $y^2$ as $2y$ instead of $2y/dydx$), or forget the product rule on mixed terms like $xy$.

Every $y$ you touch, tack on $/dydx$.

The derivative of an inverse function is the reciprocal of the derivative of the original function evaluated at the corresponding point; inverse trig derivatives follow fixed formulas.

If $g=f^-1$, then $g'(x) = /1f'(g(x))$. Also: $/ddx[ x] = /1sqrt1-x^2$, $/ddx[ x] = /11+x^2$, $/ddx[ x] = -/1sqrt1-x^2$.

$/ddx[ x]=/1sqrt1-x^2$ (positive, involves a square root) vs.\ $/ddx[ x]=/11+x^2$ (positive, no square root, denominator never zero) --- students mix up which has the radical.

Students plug $x$ directly into $/1f'(x)$ instead of $/1f'(g(x))$ when finding the derivative of an inverse function at a specific point, skipping the step of evaluating $g(x)$ first.

Inverse function derivative: flip $f'$, but evaluate at $g(x)$ first.

Position, velocity, and acceleration are linked by differentiation: velocity is the rate of change of position, and acceleration is the rate of change of velocity.

$v(t) = s'(t)$, $a(t) = v'(t) = s''(t)$. Speed is $|v(t)|$. The object speeds up when $v(t)$ and $a(t)$ have the SAME sign, and slows down when they have OPPOSITE signs.

Velocity (signed, indicates direction: positive = moving in positive direction) vs.\ speed ($|v(t)|$, always nonnegative, indicates only how fast) --- ``speeding up'' refers to speed increasing, not necessarily velocity increasing.

Students conclude an object is ``speeding up'' whenever acceleration is positive, without checking whether velocity is also positive at that instant (same-sign rule).

Same sign ($v,a$) $→$ speeding up; opposite signs $→$ slowing down.

Related rates problems connect the rates of change of two or more quantities that are linked by an equation, using implicit differentiation with respect to time.

Write an equation relating the quantities, differentiate both sides implicitly with respect to $t$ (every variable gets a $/d(·)dt$ factor via chain rule), then substitute known values AFTER differentiating.

Related rates (multiple quantities changing simultaneously with respect to TIME, requires an equation linking variables and implicit differentiation) vs.\ optimization (finding a single maximum/minimum value, requires setting $f'(x)=0$ and testing critical points, no time variable involved).

Students substitute the given numerical values into the equation BEFORE differentiating, which incorrectly turns variables into constants and produces $/ddt=0$.

Differentiate first, plug in numbers second --- never the reverse.

When a limit produces an indeterminate form $/00$ or $/∞∞$, L'Hospital's Rule allows evaluating the limit of the ratio of derivatives instead.

If $_x→ c/f(x)g(x)$ is $/00$ or $/∞∞$, then $_x→ c/f(x)g(x) = _x→ c/f'(x)g'(x)$ (provided the right-hand limit exists), reapplying as needed.

L'Hospital's Rule (differentiate numerator and denominator SEPARATELY, apply only to $/00$ or $/∞∞$) vs.\ the quotient rule (differentiates the quotient AS a single function, applies to any differentiable quotient, not just indeterminate limits).

Students apply L'Hospital's Rule to a limit that is NOT actually in $/00$ or $/∞∞$ form (forgetting to verify the indeterminate form first), or apply the quotient rule to numerator/denominator instead of differentiating each separately.

Check $/00$ or $/∞∞$ first, THEN differentiate top and bottom separately.

The differential $dy = f'(x) dx$ estimates the actual change in $y$ for a small change $dx$ in $x$, using the tangent line's slope.

$dy = f'(x) dx$ approximates $ y = f(x+dx)-f(x)$ for small $dx$. Propagated error: if $x$ has measurement error $dx$, the error in a dependent quantity is approximately $dy = f'(x) dx$.

$dy$ (the differential, the tangent-line-based approximation of the change) vs.\ $ y$ (the exact/actual change in the function) --- $dy ≈ y$ only for small $dx$, and they are generally not equal.

Students compute $ y$ exactly by evaluating $f(x+dx)-f(x)$ when the problem specifically asks for the differential approximation $dy=f'(x) dx$, or vice versa.

$dy = f'(x) dx$ $arrow$ tangent-line estimate of the true change.

A continuous function on a closed interval is guaranteed to attain an absolute maximum and minimum, and these extrema must occur at critical points or endpoints.

Extreme Value Theorem: if $f$ is continuous on $[a,b]$, $f$ attains an absolute max and min on $[a,b]$. Critical point: $x=c$ where $f'(c)=0$ or $f'(c)$ is undefined (and $c$ is in the domain of $f$). Candidates Test: evaluate $f$ at all critical points and both endpoints; largest/smallest value wins.

Critical point (where $f'(x)=0$ or undefined --- a CANDIDATE for an extremum, not automatically one) vs.\ actual local/absolute extremum (confirmed by a sign change in $f'$ or by the Candidates Test).

Students find where $f'(x)=0$ and automatically call it a maximum or minimum without testing whether $f'$ actually changes sign there (it might be neither, as in $f(x)=x^3$ at $x=0$).

Candidates Test: critical points + endpoints, biggest wins max, smallest wins min.

The Mean Value Theorem guarantees that a continuous, differentiable function achieves its average rate of change as an instantaneous rate of change somewhere on the interval.

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

MVT (requires continuity on the CLOSED interval $[a,b]$ and differentiability on the OPEN interval $(a,b)$ --- the two conditions use different interval types) vs.\ Rolle's Theorem (MVT's special case, additionally requires $f(a)=f(b)$).

Students state MVT conclusions without verifying both hypotheses (continuity on $[a,b]$ AND differentiability on $(a,b)$), losing justification credit, especially when a function has a corner inside the interval.

MVT: average slope = some instantaneous slope, guaranteed by continuity + differentiability.

The sign of $f'$ reveals where $f$ is increasing or decreasing, and sign changes in $f'$ at critical points identify local extrema.

$f$ is increasing where $f'(x)>0$ and decreasing where $f'(x)<0$. First Derivative Test: if $f'$ changes from $+$ to $-$ at $c$, $f$ has a local max at $c$; if $f'$ changes from $-$ to $+$, local min at $c$; no sign change means neither.

First Derivative Test (uses sign CHANGES of $f'$ around a critical point to classify extrema) vs.\ Second Derivative Test (uses the SIGN of $f''$ at the critical point directly --- $f''(c)>0$ means local min, $f''(c)<0$ means local max).

Students build a sign chart for $f'$ but test the wrong side of the critical point, or misread which sign change ($+→-$ vs.\ $-→+$) corresponds to a max versus a min.

$f'$: $+→-$ is a max (hill), $-→+$ is a min (valley).

The sign of $f''$ reveals the concavity of $f$, and points where concavity changes are inflection points.

$f$ is concave up where $f''(x)>0$ and concave down where $f''(x)<0$. An inflection point occurs at $x=c$ where $f''$ CHANGES SIGN (not merely where $f''(c)=0$). Second Derivative Test: if $f'(c)=0$ and $f''(c)>0$, local min; if $f''(c)<0$, local max; if $f''(c)=0$, test is inconclusive.

$f''(c)=0$ (necessary condition to CHECK for a possible inflection point) vs.\ an actual inflection point (requires $f''$ to change sign at $c$, not just equal zero there --- e.g., $f(x)=x^4$ has $f''(0)=0$ but no inflection point).

Students declare an inflection point wherever $f''(x)=0$ without verifying that $f''$ actually changes sign on either side of that $x$-value.

Inflection = $f''$ changes sign, not just $f''=0$.

The behaviors of $f$, $f'$, and $f''$ are interlocking: each derivative's sign chart constrains the shape of the graph one level down.

Given a graph of $f'$: $f$ increases where $f'>0$, has local extrema where $f'$ crosses zero, and $f$ is concave up where $f'$ is increasing (i.e., where $f'' = (f')' > 0$). Build a combined sign chart across all critical $x$-values to sketch $f$.

Reading a graph OF $f'$ to determine behavior of $f$ (increasing/decreasing, concavity) vs.\ reading a graph of $f$ itself to determine $f'$ and $f''$ --- students often analyze the wrong function's graph for the question asked.

When given the graph of $f'(x)$ (not $f(x)$), students look for where the graph ``turns around'' (a local extremum of $f'$'s graph) instead of where the graph crosses the $x$-axis, to find extrema of $f$.

Graph of $f'$: crossings of the $x$-axis = extrema of $f$; increasing sections of $f'$ = concave-up sections of $f$.

Optimization finds the absolute maximum or minimum of a quantity by expressing it as a single-variable function and applying critical-point analysis.

Write a primary equation for the quantity to optimize and a constraint equation; use the constraint to reduce the primary equation to one variable; find critical points by setting the derivative to 0; confirm the extremum via the First/Second Derivative Test or Candidates Test on the domain's endpoints.

Optimization (ONE quantity being maximized/minimized subject to a constraint, no time variable, solved via critical points) vs.\ related rates (MULTIPLE quantities changing with respect to time, solved via implicit differentiation, no maximizing/minimizing).

Students find the critical point correctly but forget to verify it's a maximum (not minimum) via a derivative test, or forget to check the domain's endpoints when the domain is a closed interval.

Optimize: one equation to optimize, one constraint to substitute in, then set derivative to zero.

A definite integral is the limit of Riemann sums --- the sum of areas of thin rectangles (or trapezoids) approximating the region under a curve.

Left Riemann sum: $_i=0^n-1 f(x_i) x$; Right Riemann sum: $_i=1^n f(x_i) x$; Midpoint sum uses midpoints of subintervals; Trapezoidal sum: $/ x2≤ft[f(x_0)+2f(x_1)+·s+2f(x_n-1)+f(x_n)]$.

Left Riemann sum overestimates on a decreasing function (and underestimates on increasing) vs.\ Right Riemann sum overestimates on an increasing function (and underestimates on decreasing) --- the over/underestimate direction flips depending on whether $f$ is increasing or decreasing.

Students misapply the over/underestimate rule by not checking BOTH monotonicity (increasing/decreasing) AND concavity together (concavity determines whether trapezoidal sums over- or under-estimate: concave up $→$ trapezoidal overestimates).

Left/Right sums: match the rectangle's corner to the curve's slope direction to see over/under.

The definite integral represents the signed accumulated area between a curve and the $x$-axis, obtained as the limit of Riemann sums as $n→∞$.

$_a^b f(x) dx = _n→∞_i=1^n f(x_i) x$. Properties: $_a^b f dx = -_b^a f dx$; $_a^a f dx = 0$; $_a^b f dx + _b^c f dx = _a^c f dx$; $_a^b [f(x)± g(x)] dx = _a^b f dx ± _a^b g dx$.

Definite integral $_a^b f(x) dx$ (a NUMBER, the signed area, possibly negative where $f<0$) vs.\ indefinite integral $ f(x) dx$ (a FAMILY OF FUNCTIONS, $F(x)+C$).

Students treat the definite integral as always representing positive area, forgetting that regions below the $x$-axis contribute negatively to $_a^b f(x) dx$ (though they add positively to total/geometric area).

Definite integral = signed area; below the axis subtracts.

The Fundamental Theorem of Calculus links differentiation and integration as inverse processes: accumulation functions can be differentiated, and definite integrals can be evaluated via antiderivatives.

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

FTC Part 1 (differentiating an integral with a variable bound, produces a FUNCTION, needs chain rule if the bound is not simply $x$) vs.\ FTC Part 2 (evaluating a definite integral using an antiderivative, produces a NUMBER).

Students forget to multiply by $g'(x)$ (the chain rule factor) when the upper limit of integration is a function of $x$ other than $x$ itself, e.g., $/ddx_0^x^2 f(t) dt = f(x^2)$ (missing the required $· 2x$).

FTC1: derivative of integral = plug in bound $×$ bound's derivative (chain rule).

Antidifferentiation reverses differentiation; because derivatives of constants vanish, every indefinite integral carries an arbitrary constant $C$.

$ 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$; $ ^2 x dx = x + C$.

$ x dx = x+C$ (no negative) vs.\ $ x dx = - x+C$ (negative sign) --- the sign appears on the sine integral, opposite of where it appears in differentiation.

Students forget ``$+C$'' on indefinite integrals (an automatic point loss on free response), or forget that $ x^-1dx = |x|+C$ is the ONE power-rule exception that doesn't follow $/x^n+1n+1$.

Antiderivative of a power: add one to exponent, divide by new exponent, plus $C$ --- always $+C$.

U-substitution reverses the chain rule, converting a composite-function integral into a simpler integral by substituting an inner expression as a new variable.

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

U-substitution on a definite integral where you CHANGE THE BOUNDS to match $u$ (then no need to substitute back) vs.\ leaving the bounds as $x$-values (then you MUST substitute $u$ back to $x$ before plugging in the original bounds) --- mixing the two approaches is a common error.

Students evaluate a definite integral using the new $u$-bounds but forget to convert them from $x$-bounds first (i.e., they plug the original $x$-limits into the antiderivative expressed in terms of $u$).

Pick $u$ = the ``inside'' function; $du$ must match what's left over exactly.

A definite integral of a rate function gives the net accumulated change in the corresponding amount function over that interval.

If $r(t)$ is a rate of change, then $_a^b r(t) dt$ = total (net) change in the quantity from $t=a$ to $t=b$: $ F(b) = F(a) + _a^b r(t) dt$. Total distance traveled = $_a^b |v(t)| dt$, while displacement = $_a^b v(t) dt$.

Displacement $_a^b v(t) dt$ (net change in position, can be negative, motion in both directions cancels) vs.\ total distance traveled $_a^b |v(t)| dt$ (always nonnegative, requires splitting the integral at sign changes of $v(t)$).

Students compute $_a^b v(t) dt$ when the question asks for total distance traveled, failing to split the interval where $v(t)$ changes sign and take absolute values on each piece.

Rate integrated over time = total change; $|v|$ integrated = total distance (split at $v=0$).

A slope field visually represents a differential equation by drawing small line segments whose slopes equal $/dydx$ at each plotted point, letting you sketch solution curves without solving algebraically.

At each point $(x,y)$, the slope field segment has slope equal to $/dydx$ evaluated at that $(x,y)$ using the given differential equation. A particular solution curve follows the segments continuously through its initial condition.

Slope field for $/dydx=f(x,y)$ (slopes may depend on BOTH $x$ and $y$, so segments vary vertically even along the same vertical line if $y$ differs) vs.\ a slope field where $/dydx=f(x)$ only (segments are identical for a fixed $x$, regardless of $y$).

Students match a slope field to a differential equation by only checking a couple of points, missing that the correct field must have zero slopes exactly where $f(x,y)=0$ and consistent signs everywhere else.

Slope field segment slope = plug the point into $/dydx$.

Euler's Method approximates a solution curve numerically by taking small linear steps, each using the slope from the differential equation at the current point.

Given $/dydx=f(x,y)$, $(x_0,y_0)$, and step size $ x$: $x_n+1=x_n+ x$ and $y_n+1=y_n+f(x_n,y_n)· x$. Repeat step by step to approximate $y$ at a target $x$-value.

Euler's Method (uses the slope at the CURRENT/left endpoint of each step, like a left-Riemann-sum-style approximation, can over/underestimate depending on concavity) vs.\ the exact solution curve (found by actually solving the differential equation, no stepwise error accumulation).

Students recompute the slope using the NEW $y$-value within the same step (i.e., they use $f(x_n+1,y_n+1)$ instead of $f(x_n,y_n)$), or forget to update BOTH $x$ and $y$ before starting the next step.

Euler: new $y$ = old $y$ + (slope at old point)$× x$, step by step.

A separable differential equation can be solved by algebraically sorting all $y$-terms to one side and $x$-terms to the other, then integrating both sides.

For $/dydx=g(x)h(y)$: separate as $/dyh(y)=g(x) dx$, integrate both sides, then solve for $y$ and use the initial condition to find $C$. Special case: $/dydx=ky ⇒ y=y_0e^kt$ (exponential growth if $k>0$, decay if $k<0$).

$/dydx=ky$ (proportional growth/decay, rate depends on the CURRENT amount, solution is exponential $y=y_0e^kt$) vs.\ $/dydx=k$ (constant rate, solution is linear $y=y_0+kt$) --- students often assume every rate problem is exponential.

Students forget to apply the initial condition to solve for $C$ (or for $y_0$) after integrating, leaving a general solution when the question asks for the PARTICULAR solution, or forget the constant of integration when first separating and integrating.

Separate: $y$'s with $dy$, $x$'s with $dx$, integrate both sides, then apply the initial condition.

The average value of a function over an interval is the constant height a rectangle of the same width would need to match the accumulated area under the curve.

Average value of $f$ on $[a,b]$: $ f_avg = /1b-a_a^b f(x) dx$. This is a direct consequence of the Mean Value Theorem for Integrals.

Average VALUE of a function $/1b-a_a^b f(x) dx$ (a single number describing typical output) vs.\ average RATE of change $/f(b)-f(a)b-a$ (slope of the secant line, used with the ORIGINAL function, not its rate) --- these use different formulas and different input functions.

Students forget the $/1b-a$ factor in front of the integral, effectively reporting the total accumulated area instead of the average height.

Average value = $/1width×$ (accumulated area).

Integration computes areas between curves and volumes of solids by summing infinitely many thin slices (rectangles for area, discs/washers/cross-sections for volume).

Area between curves: $_a^b [top(x)-bottom(x)] dx$ (or right $-$ left for horizontal slices). Disc method: $ V=_a^b [R(x)]^2 dx$. Washer method: $ V=_a^b ≤ft([R(x)]^2-[r(x)]^2)dx$. Known cross-sections: $ V=_a^b A(x) dx$, where $A(x)$ is the area formula for that cross-sectional shape (square: $s^2$; semicircle: $/12 r^2$; equilateral triangle: $/sqrt34s^2$) built on the boundary as the side/diameter.

Disc method (solid is completely filled, no hole, radius is a single function $R(x)$) vs.\ washer method (solid has a hole through it, requires OUTER radius squared MINUS inner radius squared, $[R(x)]^2 - [r(x)]^2$, not $[R(x)-r(x)]^2$).

Students subtract the radii before squaring in the washer method (computing $[R(x)-r(x)]^2$ instead of the correct $[R(x)]^2-[r(x)]^2$), or forget to subtract the axis of rotation's offset from the radius when the region does not touch the axis of revolution.

Washer: square each radius separately, THEN subtract --- never subtract first.

The handful of derivative and integral formulas below cover the overwhelming majority of AP Calculus AB computation.

Derivatives: $/ddx[x^n]=nx^n-1$; $/ddx[ x]= x$; $/ddx[ x]=- x$; $/ddx[ x]=^2 x$; $/ddx[e^x]=e^x$; $/ddx[ x]=/1x$; $/ddx[ x]=/1sqrt1-x^2$; $/ddx[ x]=/11+x^2$. Rules: $(fg)'=f'g+fg'$; $≤ft(/fg)'=/f'g-fg'g^2$; $[f(g(x))]'=f'(g(x))g'(x)$. Integrals: $ x^n dx = /x^n+1n+1+C$; $ /1xdx=|x|+C$; $ e^x dx=e^x+C$; $ x dx=- x+C$; $ x dx= x+C$. FTC: $_a^b f(x) dx=F(b)-F(a)$; $/ddx_a^x f(t) dt = f(x)$.

Power, product, quotient, chain, FTC --- the five tools behind almost every problem.

These paired terms sound similar but test opposite or unrelated ideas --- mixing them up is the single most common source of lost points.

When two terms sound alike, ask: does this involve time, a sign change, or a single point vs.\ a whole interval?

Each order of derivative answers a distinct question about the behavior of $f$; keeping the hierarchy straight prevents misreading graphs and tables.

Zeroth = where you are, first = how fast, second = how the speed is changing.

A secondary set of formulas rounds out the toolkit for geometry-based and motion-based free-response questions.

These formulas appear on nearly every FRQ set --- write them from memory before test day, not during it.

FRQs reward showing the correct calculus SETUP explicitly, not just a final numeric answer --- most points are earned for justification, not arithmetic.

Setup earns points even before the final number --- show the rule, then the substitution, then the units.

Knowing the exact structure of the exam lets you pace each section correctly and never get caught off guard by calculator restrictions.

42 MC (29 no-calc + 13 calc) + 6 FRQ (2 calc + 4 no-calc) = 3 hr 10 min, 50/50 weighting.

Most AP Calculus AB problem types follow a repeatable multi-step pathway --- recognizing which pathway applies is half the battle.

Every problem: recognize the pattern, choose the rule, execute, then interpret in context.

A large share of AP Calculus AB questions supply information as a graph or a table instead of an explicit formula --- misreading which function is displayed is the top non-computational error.

Name the function you're looking at first (is it $f$, $f'$, or $f''$?), then read its features.

These statements sound like valid calculus rules but are false or incomplete as written --- recognizing them prevents costly exam errors.

When a rule ``sounds right,'' check its hypotheses before trusting its conclusion.