Power rule d d x x n = n x n − 1 \dfrac{d}{dx} x^n = n x^{n-1} d x d x n = n x n − 1 for any real exponent n n n .Chain rule d d x f ( g ( x ) ) = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \dfrac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) d x d f ( g ( x )) = f ′ ( g ( x )) ⋅ g ′ ( x ) — outer derivative at the inner, times the inner derivative.Fundamental Theorem of Calculus (Part 2) ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) , where F F F is any antiderivative of f f f .Definition of the derivative f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h f'(a) = \lim_{h \to 0} \dfrac{f(a+h) - f(a)}{h} f ′ ( a ) = lim h → 0 h f ( a + h ) − f ( a ) — the slope of the tangent line at x = a x = a x = a .Mean Value Theorem If f f f is continuous on [ a , b ] [a,b] [ a , b ] and differentiable on ( a , b ) (a,b) ( a , b ) , some c c c gives f ′ ( c ) = f ( b ) − f ( a ) b − a f'(c) = \dfrac{f(b) - f(a)}{b - a} f ′ ( c ) = b − a f ( b ) − f ( a ) . Definition of a limit The single value f ( x ) f(x) f ( x ) approaches as x → a x \to a x → a . It can exist even when f ( a ) f(a) f ( a ) is undefined. Conditions for continuity at x = a (1) f ( a ) f(a) f ( a ) is defined, (2) lim x → a f ( x ) \lim_{x \to a} f(x) lim x → a f ( x ) exists, and (3) the limit equals f ( a ) f(a) f ( a ) . Two-sided limit exists when… The left-hand and right-hand limits are equal: lim x → a − f ( x ) = lim x → a + f ( x ) \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) lim x → a − f ( x ) = lim x → a + f ( x ) . Limit: sin(x)/x as x → 0 lim x → 0 sin x x = 1 \lim_{x \to 0} \dfrac{\sin x}{x} = 1 lim x → 0 x sin x = 1 — a key special limit.Limit: (1 − cos x)/x as x → 0 lim x → 0 1 − cos x x = 0 \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0 lim x → 0 x 1 − cos x = 0 .Removable discontinuity A 'hole': the limit exists but doesn't equal f ( a ) f(a) f ( a ) (or f ( a ) f(a) f ( a ) is undefined). Often from a factor that cancels.
Jump discontinuity The left- and right-hand limits both exist but disagree, so the two-sided limit does not exist. Infinite discontinuity The function approaches ± ∞ \pm\infty ± ∞ at the point — a vertical asymptote. Squeeze theorem If g ( x ) ≤ f ( x ) ≤ h ( x ) g(x) \le f(x) \le h(x) g ( x ) ≤ f ( x ) ≤ h ( x ) near a a a and lim g = lim h = L \lim g = \lim h = L lim g = lim h = L , then lim f = L \lim f = L lim f = L . Limit at infinity of a rational function Compare degrees: lower numerator → 0; equal → ratio of leading coefficients; higher numerator → ± ∞ \pm\infty ± ∞ . Intermediate Value Theorem (IVT) If f f f is continuous on [ a , b ] [a,b] [ a , b ] , it takes every value between f ( a ) f(a) f ( a ) and f ( b ) f(b) f ( b ) ; opposite signs force a root between.
When is a function NOT continuous? When any of the three conditions fails — at a removable (hole), jump, or infinite discontinuity.
Evaluating a 0/0 limit Use algebra first — factor and cancel, multiply by a conjugate, or simplify a complex fraction — then substitute. Horizontal asymptote and limits y = L y = L y = L is a horizontal asymptote if lim x → ∞ f ( x ) = L \lim_{x \to \infty} f(x) = L lim x → ∞ f ( x ) = L or lim x → − ∞ f ( x ) = L \lim_{x \to -\infty} f(x) = L lim x → − ∞ f ( x ) = L .Vertical asymptote and limits x = a x = a x = a is a vertical asymptote if f ( x ) → ± ∞ f(x) \to \pm\infty f ( x ) → ± ∞ as x → a x \to a x → a from one or both sides.Direct substitution property If f f f is continuous at a a a , then lim x → a f ( x ) = f ( a ) \lim_{x \to a} f(x) = f(a) lim x → a f ( x ) = f ( a ) — just plug in. Limit of a constant lim x → a c = c \lim_{x \to a} c = c lim x → a c = c — a constant function's limit is the constant.Continuity of polynomials Every polynomial is continuous for all real x x x ; rational functions are continuous except where the denominator is 0. End behavior from leading term For large ∣ x ∣ |x| ∣ x ∣ , a polynomial behaves like its leading term, which sets the limits at ± ∞ \pm\infty ± ∞ . Limit: e x e^x e x as x → − ∞ x \to -\infty x → − ∞ lim x → − ∞ e x = 0 \lim_{x \to -\infty} e^x = 0 lim x → − ∞ e x = 0 ; and lim x → ∞ e x = ∞ \lim_{x \to \infty} e^x = \infty lim x → ∞ e x = ∞ .
Continuity of a piecewise function Continuous at the boundary if the two pieces' one-sided limits match the defined value there. Existence of a limit vs. value of f A limit describes where the function is headed; it can exist even where f f f is undefined or differs.
Limit of a sum/product/quotient Limits distribute over sums, products, and quotients (denominator limit ≠ 0), so split a complex limit into parts. Difference quotient f ( a + h ) − f ( a ) h \dfrac{f(a+h) - f(a)}{h} h f ( a + h ) − f ( a ) — the average rate of change (a secant slope). Its limit is the derivative.Geometric meaning of f′(a) The slope of the line tangent to the graph of f f f at x = a x = a x = a — the instantaneous rate of change. Constant rule d d x c = 0 \dfrac{d}{dx} c = 0 d x d c = 0 — the derivative of a constant is zero.Constant-multiple rule d d x [ c f ( x ) ] = c f ′ ( x ) \dfrac{d}{dx} [c f(x)] = c f'(x) d x d [ c f ( x )] = c f ′ ( x ) .Sum/difference rule ( f ± g ) ′ = f ′ ± g ′ (f \pm g)' = f' \pm g' ( f ± g ) ′ = f ′ ± g ′ — differentiate term by term.Product rule ( f g ) ′ = f ′ g + f g ′ (f g)' = f' g + f g' ( f g ) ′ = f ′ g + f g ′ .Quotient rule ( f g ) ′ = f ′ g − f g ′ g 2 \left( \dfrac{f}{g} \right)' = \dfrac{f' g - f g'}{g^2} ( g f ) ′ = g 2 f ′ g − f g ′ — 'low d-high minus high d-low, over low squared.'Derivative of e x e^x e x d d x e x = e x \dfrac{d}{dx} e^x = e^x d x d e x = e x — its own derivative.Derivative of ln x d d x ln x = 1 x \dfrac{d}{dx} \ln x = \dfrac{1}{x} d x d ln x = x 1 (for x > 0 x > 0 x > 0 ).Derivative of sin x d d x sin x = cos x \dfrac{d}{dx} \sin x = \cos x d x d sin x = cos x .Derivative of cos x d d x cos x = − sin x \dfrac{d}{dx} \cos x = -\sin x d x d cos x = − sin x .Derivative of tan x d d x tan x = sec 2 x \dfrac{d}{dx} \tan x = \sec^2 x d x d tan x = sec 2 x .Derivative of sec x d d x sec x = sec x tan x \dfrac{d}{dx} \sec x = \sec x \tan x d x d sec x = sec x tan x .Derivative of cot x d d x cot x = − csc 2 x \dfrac{d}{dx} \cot x = -\csc^2 x d x d cot x = − csc 2 x .Derivative of csc x d d x csc x = − csc x cot x \dfrac{d}{dx} \csc x = -\csc x \cot x d x d csc x = − csc x cot x .Derivative of a x a^x a x d d x a x = a x ln a \dfrac{d}{dx} a^x = a^x \ln a d x d a x = a x ln a for a constant base a > 0 a > 0 a > 0 .Second derivative f ′ ′ ( x ) f''(x) f ′′ ( x ) , the derivative of f ′ f' f ′ — measures concavity and is acceleration in motion problems.Differentiability implies continuity If f f f is differentiable at a a a , it is continuous at a a a . The converse is false. Where derivatives fail to exist At a corner, a cusp, a vertical tangent, or any discontinuity — even if f f f is continuous there. Rewrite before the power rule Convert roots and reciprocals to powers first: x = x 1 / 2 \sqrt{x} = x^{1/2} x = x 1/2 , 1 x 2 = x − 2 \dfrac{1}{x^2} = x^{-2} x 2 1 = x − 2 . Derivative of a linear function d d x ( m x + b ) = m \dfrac{d}{dx}(mx + b) = m d x d ( m x + b ) = m — the slope is constant.Estimating f′ from a table Use the difference quotient of two nearby points: f ( x 2 ) − f ( x 1 ) x 2 − x 1 \dfrac{f(x_2) - f(x_1)}{x_2 - x_1} x 2 − x 1 f ( x 2 ) − f ( x 1 ) . Tangent line equation at x = a y = f ( a ) + f ′ ( a ) ( x − a ) y = f(a) + f'(a)(x - a) y = f ( a ) + f ′ ( a ) ( x − a ) — point-slope form using the derivative as slope.Normal line The line perpendicular to the tangent at a point; its slope is − 1 f ′ ( a ) -\dfrac{1}{f'(a)} − f ′ ( a ) 1 . Higher-order derivatives notation f ′ ( x ) , f ′ ′ ( x ) , f ′ ′ ′ ( x ) , f ( 4 ) ( x ) , … f'(x), f''(x), f'''(x), f^{(4)}(x), \dots f ′ ( x ) , f ′′ ( x ) , f ′′′ ( x ) , f ( 4 ) ( x ) , … — repeatedly differentiate.Derivative of √x d d x x = 1 2 x \dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2\sqrt{x}} d x d x = 2 x 1 .Derivative of 1/x d d x 1 x = − 1 x 2 \dfrac{d}{dx} \dfrac{1}{x} = -\dfrac{1}{x^2} d x d x 1 = − x 2 1 .Average vs. instantaneous rate of change Average = secant slope over [ a , b ] [a,b] [ a , b ] ; instantaneous = tangent slope = derivative at a point.
Chain rule in words Derivative of the outside (leaving the inside alone) times the derivative of the inside. d/dx sin(g(x)) cos ( g ( x ) ) ⋅ g ′ ( x ) \cos(g(x)) \cdot g'(x) cos ( g ( x )) ⋅ g ′ ( x ) — e.g. d d x sin ( 4 x ) = 4 cos ( 4 x ) \dfrac{d}{dx}\sin(4x) = 4\cos(4x) d x d sin ( 4 x ) = 4 cos ( 4 x ) .d d x e g ( x ) \dfrac{d}{dx} e^{g(x)} d x d e g ( x ) e g ( x ) ⋅ g ′ ( x ) e^{g(x)} \cdot g'(x) e g ( x ) ⋅ g ′ ( x ) — e.g. d d x e 3 x 2 = 6 x e 3 x 2 \dfrac{d}{dx} e^{3x^2} = 6x\,e^{3x^2} d x d e 3 x 2 = 6 x e 3 x 2 .d/dx ln(g(x)) g ′ ( x ) g ( x ) \dfrac{g'(x)}{g(x)} g ( x ) g ′ ( x ) — the inner derivative over the inner function.d d x [ g ( x ) ] n \dfrac{d}{dx} [g(x)]^n d x d [ g ( x ) ] n n [ g ( x ) ] n − 1 ⋅ g ′ ( x ) n[g(x)]^{n-1} \cdot g'(x) n [ g ( x ) ] n − 1 ⋅ g ′ ( x ) — the power rule with the chain rule.Implicit differentiation Differentiate both sides in x x x , attaching d y d x \dfrac{dy}{dx} d x d y to every y y y -term, then solve for d y d x \dfrac{dy}{dx} d x d y . Why y-terms get dy/dx Because y y y is treated as a function of x x x , so the chain rule applies whenever you differentiate a y y y -term. Derivative of an inverse function If g = f − 1 g = f^{-1} g = f − 1 , then g ′ ( x ) = 1 f ′ ( g ( x ) ) g'(x) = \dfrac{1}{f'(g(x))} g ′ ( x ) = f ′ ( g ( x )) 1 . Derivative of arcsin x d d x arcsin x = 1 1 − x 2 \dfrac{d}{dx} \arcsin x = \dfrac{1}{\sqrt{1 - x^2}} d x d arcsin x = 1 − x 2 1 .Derivative of arccos x d d x arccos x = − 1 1 − x 2 \dfrac{d}{dx} \arccos x = -\dfrac{1}{\sqrt{1 - x^2}} d x d arccos x = − 1 − x 2 1 .Derivative of arctan x d d x arctan x = 1 1 + x 2 \dfrac{d}{dx} \arctan x = \dfrac{1}{1 + x^2} d x d arctan x = 1 + x 2 1 .d/dx cos(x²) − 2 x sin ( x 2 ) -2x \sin(x^2) − 2 x sin ( x 2 ) by the chain rule (outer − sin -\sin − sin , inner 2 x 2x 2 x ).d/dx √(x² + 1) x x 2 + 1 \dfrac{x}{\sqrt{x^2 + 1}} x 2 + 1 x — chain rule on ( x 2 + 1 ) 1 / 2 (x^2+1)^{1/2} ( x 2 + 1 ) 1/2 .d/dx tan(x³) 3 x 2 sec 2 ( x 3 ) 3x^2 \sec^2(x^3) 3 x 2 sec 2 ( x 3 ) by the chain rule.d d x ( 2 x + 1 ) 5 \dfrac{d}{dx} (2x + 1)^5 d x d ( 2 x + 1 ) 5 10 ( 2 x + 1 ) 4 10(2x + 1)^4 10 ( 2 x + 1 ) 4 — bring down 5, reduce power, times inner derivative 2.Implicit slope at a point Find d y d x \dfrac{dy}{dx} d x d y implicitly, then substitute the point's coordinates to get the tangent slope. Nesting the chain rule For f ( g ( h ( x ) ) ) f(g(h(x))) f ( g ( h ( x ))) , differentiate outermost to innermost, multiplying all the inner derivatives. Logarithmic differentiation Take ln \ln ln of both sides before differentiating — useful for products, quotients, or variable exponents. d/dx ln(5x) 1 x \dfrac{1}{x} x 1 — since 5 5 x = 1 x \dfrac{5}{5x} = \dfrac{1}{x} 5 x 5 = x 1 .Chain rule with a table If h ( x ) = f ( g ( x ) ) h(x) = f(g(x)) h ( x ) = f ( g ( x )) , then h ′ ( x ) = f ′ ( g ( x ) ) g ′ ( x ) h'(x) = f'(g(x)) g'(x) h ′ ( x ) = f ′ ( g ( x )) g ′ ( x ) ; read the needed values from the table.
Tangent to an implicit curve Differentiate implicitly, substitute the point for the slope, then write the tangent line in point-slope form. d d x a g ( x ) \dfrac{d}{dx} a^{g(x)} d x d a g ( x ) a g ( x ) ln a ⋅ g ′ ( x ) a^{g(x)} \ln a \cdot g'(x) a g ( x ) ln a ⋅ g ′ ( x ) — exponential chain rule with base a a a .Related rates: the idea Two quantities linked by an equation change with time; differentiate the equation in t t t and solve for the unknown rate. Related rates: order of steps Write the relating equation, differentiate with respect to t t t , THEN substitute the known values and rates. Linearization (tangent-line approx.) L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L(x) = f(a) + f'(a)(x - a) L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) estimates f f f near a a a .When can you use L'Hopital's Rule? Only when the limit is the indeterminate form 0 0 \dfrac{0}{0} 0 0 or ∞ ∞ \dfrac{\infty}{\infty} ∞ ∞ — verify first. L'Hopital's Rule For an indeterminate 0 0 \dfrac{0}{0} 0 0 or ∞ ∞ \dfrac{\infty}{\infty} ∞ ∞ , lim f g = lim f ′ g ′ \lim \dfrac{f}{g} = \lim \dfrac{f'}{g'} lim g f = lim g ′ f ′ . Velocity from position v ( t ) = s ′ ( t ) v(t) = s'(t) v ( t ) = s ′ ( t ) — velocity is the derivative of position.Acceleration from position a ( t ) = v ′ ( t ) = s ′ ′ ( t ) a(t) = v'(t) = s''(t) a ( t ) = v ′ ( t ) = s ′′ ( t ) — acceleration is the second derivative of position.Particle at rest The particle is at rest when v ( t ) = 0 v(t) = 0 v ( t ) = 0 ; these are candidate times for a change of direction. Speeding up vs. slowing down Speeding up when v v v and a a a have the SAME sign; slowing down when their signs DIFFER. Speed vs. velocity Speed is ∣ v ( t ) ∣ |v(t)| ∣ v ( t ) ∣ (always ≥ 0); velocity carries a sign giving direction. Change of direction Occurs where v ( t ) v(t) v ( t ) changes sign (passes through 0), not merely where v = 0 v = 0 v = 0 . Related rates: expanding circle From A = π r 2 A = \pi r^2 A = π r 2 , d A d t = 2 π r d r d t \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt} d t d A = 2 π r d t d r . Related rates: sphere volume From V = 4 3 π r 3 V = \tfrac{4}{3}\pi r^3 V = 3 4 π r 3 , d V d t = 4 π r 2 d r d t \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt} d t d V = 4 π r 2 d t d r . Related rates: cube volume From V = s 3 V = s^3 V = s 3 , d V d t = 3 s 2 d s d t \dfrac{dV}{dt} = 3s^2 \dfrac{ds}{dt} d t d V = 3 s 2 d t d s . Ladder-against-wall setup Use x 2 + y 2 = L 2 x^2 + y^2 = L^2 x 2 + y 2 = L 2 ; differentiate to get 2 x d x d t + 2 y d y d t = 0 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0 2 x d t d x + 2 y d t d y = 0 .
L'Hopital common trap Applying it to a limit that ISN'T indeterminate (e.g. a 0/3 form) gives a wrong answer. Differential dy d y = f ′ ( x ) d x dy = f'(x)\,dx d y = f ′ ( x ) d x — the change along the tangent line, used for linear approximation.
Approximation over/underestimate A tangent-line estimate overestimates when the graph is concave down and underestimates when concave up. Total distance vs. displacement (rates) Net change uses signed velocity; total distance uses ∣ v ∣ |v| ∣ v ∣ (split where v v v changes sign).
Rate of change as a derivative Any 'how fast is __ changing' question is a derivative with respect to time. Critical point A point where f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 or f ′ ( x ) f'(x) f ′ ( x ) is undefined — a candidate for a relative extremum. f′ > 0 means… f f f is increasing on that interval (graph rising).f′ < 0 means… f f f is decreasing on that interval (graph falling).First Derivative Test At a critical point: f ′ f' f ′ changing + → − + \to - + → − is a relative max; − → + - \to + − → + is a relative min. Second Derivative Test At a critical point: f ′ ′ > 0 f'' > 0 f ′′ > 0 ⟹ relative min; f ′ ′ < 0 f'' < 0 f ′′ < 0 ⟹ relative max; f ′ ′ = 0 f'' = 0 f ′′ = 0 ⟹ inconclusive. Concave up Where f ′ ′ ( x ) > 0 f''(x) > 0 f ′′ ( x ) > 0 — the graph cups upward (∪); f ′ f' f ′ is increasing. Concave down Where f ′ ′ ( x ) < 0 f''(x) < 0 f ′′ ( x ) < 0 — the graph caps downward (∩); f ′ f' f ′ is decreasing. Inflection point A point where concavity changes — where f ′ ′ ( x ) f''(x) f ′′ ( x ) changes sign.
Extreme Value Theorem (EVT) A function continuous on a closed, bounded interval attains both an absolute max and an absolute min there. Finding absolute extrema on [a,b] Evaluate f f f at all critical points AND both endpoints; the largest/smallest values are the absolute extrema. MVT hypotheses f f f continuous on the closed [ a , b ] [a,b] [ a , b ] AND differentiable on the open ( a , b ) (a,b) ( a , b ) .Rolle's Theorem A special MVT: if also f ( a ) = f ( b ) f(a) = f(b) f ( a ) = f ( b ) , then some c c c in ( a , b ) (a,b) ( a , b ) has f ′ ( c ) = 0 f'(c) = 0 f ′ ( c ) = 0 .
Optimization steps Express the quantity as a function of one variable, differentiate, find critical points, and test them (and endpoints). When Second Derivative Test fails If f ′ ′ ( c ) = 0 f''(c) = 0 f ′′ ( c ) = 0 at the critical point, it's inconclusive — use the First Derivative Test instead.
Relative (local) extremum A point higher (max) or lower (min) than all nearby points; occurs only at critical points. Critical point that isn't an extremum Possible — e.g. f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 has f ′ ( 0 ) = 0 f'(0) = 0 f ′ ( 0 ) = 0 but no extremum there (sign of f ′ f' f ′ doesn't change). Increasing/decreasing from f′ Sign-analyze f ′ f' f ′ : positive intervals are increasing, negative intervals are decreasing. Concavity from f″ Sign-analyze f ′ ′ f'' f ′′ : positive intervals are concave up, negative intervals are concave down. f′ and the graph of f Where f ′ f' f ′ is zero/undefined are extrema candidates; where f ′ f' f ′ is large the graph is steep.
Candidates test For absolute extrema on a closed interval, the only candidates are critical points and endpoints. Connecting f, f′, f″ graphically A max of f f f sits where f ′ = 0 f' = 0 f ′ = 0 and f ′ ′ < 0 f'' < 0 f ′′ < 0 ; an inflection of f f f sits where f ′ ′ = 0 f'' = 0 f ′′ = 0 changes sign.
MVT interpretation Somewhere the instantaneous rate (tangent slope) equals the average rate (secant slope) over the interval. Definite integral meaning ∫ a b f ( x ) d x \int_a^b f(x)\,dx ∫ a b f ( x ) d x is the signed area between f f f and the x x x -axis from a a a to b b b .
Riemann sum An area approximation using rectangles whose heights come from the function at left, right, or midpoints.
Left vs. right Riemann sum (increasing f) For an increasing function, the left sum underestimates and the right sum overestimates the true area. FTC Part 1 d d x ∫ a x f ( t ) d t = f ( x ) \dfrac{d}{dx} \int_a^x f(t)\,dt = f(x) d x d ∫ a x f ( t ) d t = f ( x ) — differentiating an accumulation function returns the integrand.FTC Part 2 (evaluation) ∫ a b f ( x ) d x = F ( b ) − F ( a ) \int_a^b f(x)\,dx = F(b) - F(a) ∫ a b f ( x ) d x = F ( b ) − F ( a ) , where F F F is any antiderivative of f f f .Antiderivative A function F F F with F ′ = f F' = f F ′ = f . The indefinite integral ∫ f d x = F ( x ) + C \int f\,dx = F(x) + C ∫ f d x = F ( x ) + C . Power rule for integrals ∫ x n d x = x n + 1 n + 1 + C \int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C ∫ x n d x = n + 1 x n + 1 + C for n ≠ − 1 n \ne -1 n = − 1 .∫ 1/x dx ∫ 1 x d x = ln ∣ x ∣ + C \int \dfrac{1}{x}\,dx = \ln|x| + C ∫ x 1 d x = ln ∣ x ∣ + C .∫ e x d x \int e^x\,dx ∫ e x d x ∫ e x d x = e x + C \int e^x\,dx = e^x + C ∫ e x d x = e x + C .∫ cos x dx ∫ cos x d x = sin x + C \int \cos x\,dx = \sin x + C ∫ cos x d x = sin x + C .∫ sin x dx ∫ sin x d x = − cos x + C \int \sin x\,dx = -\cos x + C ∫ sin x d x = − cos x + C .∫ sec² x dx ∫ sec 2 x d x = tan x + C \int \sec^2 x\,dx = \tan x + C ∫ sec 2 x d x = tan x + C .u-substitution Reverse the chain rule: let u u u be an inner function, compute d u du d u , and rewrite the integral in u u u . Definite integral with substitution Either change the limits to u u u -values or convert back to x x x before evaluating. Accumulation function g ( x ) = ∫ a x f ( t ) d t g(x) = \int_a^x f(t)\,dt g ( x ) = ∫ a x f ( t ) d t stores total change; by FTC Part 1, g ′ ( x ) = f ( x ) g'(x) = f(x) g ′ ( x ) = f ( x ) .Net change theorem ∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) \int_a^b f'(x)\,dx = f(b) - f(a) ∫ a b f ′ ( x ) d x = f ( b ) − f ( a ) — the integral of a rate gives the net change.Properties: reversing limits ∫ a b f d x = − ∫ b a f d x \int_a^b f\,dx = -\int_b^a f\,dx ∫ a b f d x = − ∫ b a f d x , and ∫ a a f d x = 0 \int_a^a f\,dx = 0 ∫ a a f d x = 0 .Additivity of integrals ∫ a c f d x = ∫ a b f d x + ∫ b c f d x \int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx ∫ a c f d x = ∫ a b f d x + ∫ b c f d x .Constant multiple in an integral ∫ c f ( x ) d x = c ∫ f ( x ) d x \int c f(x)\,dx = c \int f(x)\,dx ∫ c f ( x ) d x = c ∫ f ( x ) d x — pull constants out.
Midpoint Riemann sum Uses the function value at each subinterval's midpoint; often more accurate than left or right sums.
Trapezoidal approximation Averages left and right sums; for a concave-up function it overestimates the integral. Signed area below the axis Area where f < 0 f < 0 f < 0 counts as negative in a definite integral. ∫ a x d x \int a^x\,dx ∫ a x d x ∫ a x d x = a x ln a + C \int a^x\,dx = \dfrac{a^x}{\ln a} + C ∫ a x d x = ln a a x + C for base a > 0 , a ≠ 1 a > 0,\ a \ne 1 a > 0 , a = 1 .∫ 1/(1+x²) dx ∫ 1 1 + x 2 d x = arctan x + C \int \dfrac{1}{1 + x^2}\,dx = \arctan x + C ∫ 1 + x 2 1 d x = arctan x + C .FTC with a variable upper limit g(x) d d x ∫ a g ( x ) f ( t ) d t = f ( g ( x ) ) ⋅ g ′ ( x ) \dfrac{d}{dx} \int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x) d x d ∫ a g ( x ) f ( t ) d t = f ( g ( x )) ⋅ g ′ ( x ) — chain rule on the limit.Why +C on indefinite integrals Antiderivatives differ by a constant, so the general antiderivative includes an arbitrary + C + C + C . Definite integral evaluates to a number Unlike an indefinite integral (a family of functions), ∫ a b f d x \int_a^b f\,dx ∫ a b f d x is a single value. Riemann sum → integral As the number of subintervals → ∞ (width → 0), the Riemann sum approaches ∫ a b f d x \int_a^b f\,dx ∫ a b f d x . Differential equation An equation relating a function to its derivative(s), e.g. d y d x = k y \dfrac{dy}{dx} = ky d x d y = k y . Separable differential equation Can be rewritten so all y y y -terms (with d y dy d y ) are on one side and all x x x -terms (with d x dx d x ) on the other. Solving a separable equation Separate variables, integrate both sides (add C C C ), then use an initial condition to find C C C . Exponential growth/decay model d y d t = k y \dfrac{dy}{dt} = ky d t d y = k y has solution y = y 0 e k t y = y_0 e^{kt} y = y 0 e k t ; growth if k > 0 k > 0 k > 0 , decay if k < 0 k < 0 k < 0 .Slope field A grid of short segments with slope d y d x = f ( x , y ) \dfrac{dy}{dx} = f(x,y) d x d y = f ( x , y ) at each point, picturing solution curves. Reading a slope field A solution curve follows the nearby segment directions; segments are horizontal where d y d x = 0 \dfrac{dy}{dx} = 0 d x d y = 0 . General vs. particular solution The general solution has the + C +C + C ; a particular solution fixes C C C using an initial condition. Initial condition A given point y ( x 0 ) = y 0 y(x_0) = y_0 y ( x 0 ) = y 0 used to solve for the constant of integration. Doubling time / half-life From y = y 0 e k t y = y_0 e^{kt} y = y 0 e k t : doubling time solves 2 = e k t 2 = e^{kt} 2 = e k t ; half-life solves 1 2 = e k t \tfrac{1}{2} = e^{kt} 2 1 = e k t . Sign of the square root in a solution When solving for y y y from y 2 = … y^2 = \dots y 2 = … , the initial condition decides whether to take the + or − root. Slope field: where segments are steep Where ∣ f ( x , y ) ∣ |f(x,y)| ∣ f ( x , y ) ∣ is large; horizontal where f ( x , y ) = 0 f(x,y) = 0 f ( x , y ) = 0 and vertical-ish where it's undefined. Verifying a solution Substitute y y y and its derivative back into the differential equation; both sides must match.
dP/dt = kP describes… A quantity whose rate of change is proportional to its current amount — exponential growth or decay.
Solution curves don't cross In a slope field, distinct solution curves never intersect (each point has one slope). Separating dy/dx = g(x)h(y) Rewrite as 1 h ( y ) d y = g ( x ) d x \dfrac{1}{h(y)}\,dy = g(x)\,dx h ( y ) 1 d y = g ( x ) d x , then integrate both sides.
Why AB only solves separable DEs Analytic solution methods on AP Calculus AB are limited to separable first-order equations. Area between two curves ∫ a b [ f ( x ) − g ( x ) ] d x \int_a^b [f(x) - g(x)]\,dx ∫ a b [ f ( x ) − g ( x )] d x where f f f is the top curve and g g g the bottom — always top minus bottom.Setting limits for area between curves Find the intersection points; they are the limits a a a and b b b of integration. Average value of a function 1 b − a ∫ a b f ( x ) d x \dfrac{1}{b - a} \int_a^b f(x)\,dx b − a 1 ∫ a b f ( x ) d x — the constant height of an equal-area rectangle on [ a , b ] [a,b] [ a , b ] .Displacement of a particle ∫ t 1 t 2 v ( t ) d t \int_{t_1}^{t_2} v(t)\,dt ∫ t 1 t 2 v ( t ) d t — the net change in position (signed).Total distance traveled ∫ t 1 t 2 ∣ v ( t ) ∣ d t \int_{t_1}^{t_2} |v(t)|\,dt ∫ t 1 t 2 ∣ v ( t ) ∣ d t — split the integral where the velocity changes sign.Disk method V = π ∫ R 2 d x V = \pi \int R^2\,dx V = π ∫ R 2 d x — use when the region touches the axis of rotation (no gap).Washer method V = π ∫ ( R 2 − r 2 ) d x V = \pi \int (R^2 - r^2)\,dx V = π ∫ ( R 2 − r 2 ) d x — use when a gap leaves a hollow core; R R R outer, r r r inner radius.Known cross sections V = ∫ A ( x ) d x V = \int A(x)\,dx V = ∫ A ( x ) d x — integrate the area of one cross section (square, triangle, semicircle) along the axis.Square cross sections on a base If the side length is s ( x ) s(x) s ( x ) , the cross-section area is s ( x ) 2 s(x)^2 s ( x ) 2 ; integrate that along the base. Accumulated change from a rate ∫ a b R ( t ) d t \int_a^b R(t)\,dt ∫ a b R ( t ) d t gives the total accumulated amount when R R R is a rate of change.Position from velocity + initial value s ( t ) = s ( t 0 ) + ∫ t 0 t v ( τ ) d τ s(t) = s(t_0) + \int_{t_0}^{t} v(\tau)\,d\tau s ( t ) = s ( t 0 ) + ∫ t 0 t v ( τ ) d τ — add the net change to the starting position.
Top-minus-bottom sign error Subtracting bottom from top reversed flips the sign — area must be positive, so order matters.
Curves that cross within [a,b] Split the integral at each crossing so 'top minus bottom' stays correct on each piece. Average value vs. average rate of change Average VALUE integrates the function; average RATE OF CHANGE is the secant slope f ( b ) − f ( a ) b − a \dfrac{f(b)-f(a)}{b-a} b − a f ( b ) − f ( a ) . Disk radius from the function When revolving y = f ( x ) y = f(x) y = f ( x ) about the x x x -axis, the disk radius is R = f ( x ) R = f(x) R = f ( x ) , so area = π f ( x ) 2 = \pi f(x)^2 = π f ( x ) 2 . Integrating with respect to y Revolve about the y y y -axis (or integrate vertical extent) by writing radii as functions of y y y and ∫ … d y \int \dots dy ∫ … d y . Mean Value Theorem for Integrals A continuous f f f attains its average value at some c c c in [ a , b ] [a,b] [ a , b ] .
Area as a limit of Riemann sums Area between curves is the limit of summing thin vertical strips of height (top − bottom).
Tangent-line approximation accuracy Best near the point of tangency; error grows as you move away and as curvature increases.
f′(x) = 0 vs. f′(x) undefined Both give critical points; the second includes corners, cusps, and vertical tangents.
Continuity needed for the IVT/EVT/MVT/FTC Continuity is a required hypothesis for all four major theorems — always check it. Antiderivative of a constant k ∫ k d x = k x + C \int k\,dx = kx + C ∫ k d x = k x + C .Derivative of a constant times a function Constants factor straight through: ( c f ) ′ = c f ′ (cf)' = c f' ( c f ) ′ = c f ′ and ∫ c f d x = c ∫ f d x \int cf\,dx = c\int f\,dx ∫ c f d x = c ∫ f d x . Implicit second derivative Differentiate d y d x \dfrac{dy}{dx} d x d y again implicitly, substituting the first derivative where d y d x \dfrac{dy}{dx} d x d y appears.
Position, velocity, acceleration chain Differentiate position → velocity → acceleration; integrate acceleration → velocity → position.
Global vs. local extrema Local extrema are highest/lowest nearby; global (absolute) extrema are highest/lowest over the whole interval. u-sub example: ∫ 2x(x²+1)⁴ dx u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 , d u = 2 x d x du = 2x\,dx d u = 2 x d x ⟹ ∫ u 4 d u = 1 5 ( x 2 + 1 ) 5 + C \int u^4\,du = \tfrac{1}{5}(x^2+1)^5 + C ∫ u 4 d u = 5 1 ( x 2 + 1 ) 5 + C .Limit definition at a general x f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} f ′ ( x ) = lim h → 0 h f ( x + h ) − f ( x ) gives the derivative as a function.
Concavity and the tangent line A concave-up graph lies above its tangent lines; a concave-down graph lies below them. Definite integral of an odd function over [−a, a] Equals 0 by symmetry; an even function gives 2 ∫ 0 a f d x 2\int_0^a f\,dx 2 ∫ 0 a f d x . Average rate vs. MVT The MVT guarantees the instantaneous rate equals the average rate at some interior point c c c . Estimating area from a graph/table Use left, right, midpoint, or trapezoidal sums with the given values to approximate ∫ f d x \int f\,dx ∫ f d x . Particle returns to start When displacement ∫ v d t = 0 \int v\,dt = 0 ∫ v d t = 0 over the interval, the particle ends where it began. Derivative as a rate in context Units of f ′ ( x ) f'(x) f ′ ( x ) are units of f f f per unit of x x x (e.g. meters per second). Integral as accumulation in context Units of ∫ f d x \int f\,dx ∫ f d x are units of f f f times units of x x x (e.g. gallons when f f f is gal/min). Sketching f from f′ Use f ′ f' f ′ sign for increase/decrease, f ′ = 0 f' = 0 f ′ = 0 for extrema, and f ′ ′ f'' f ′′ sign for concavity to draw f f f .