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AP Calculus AB Formulas & Justifications Limits at Infinity: To find lim f ( x) think 1 2 x Top Heavy limit is ±∞ Bottom Heavy limit is 0 Equal limit is ratio of coefficients Limits with Infinity (at vertical asymptotes): When finding a one-sided limit at a vertical asymptote, the answer is either ±∞. Justifying that a function is continuous at a point: f is continuous at c iff: 1. f (c ) is defined 3 2. lim f ( x ) exists x c 3. f (c ) = lim f ( x ) x c 4 Definition of the Derivative: f ( x h) f ( x ) f ( x x) f ( x) f '( x) lim lim h 0 x 0 h x f ( x) f (a) (Alternate form for a derivative at a given value.) xa Justifying that a derivative exists at a point, c : Show algebraically that lim f '( x) lim f '( x) . f '(a) lim x a 5 x c 6 7 8 9 10 x c Average Rate of Change of f on [a, b]: f (b) f (a ) y a.r.c. = (algebra slope of ) ba x Instantaneous Rate of Change of f at a : f '(a) (derivative at the given value) Power Rule: d n x nx n 1 dx Common Derivatives to Remember: d 1 1 dx x x 2 Trig Function Derivatives: d sin x cos x dx d tan x sec2 x dx d sec x sec x tan x dx AB Calculus Formulas & Justifications (slope of secant line) (slope of tangent line) d 1 x dx 2 x d cos x sin x dx d cot x csc2 x dx d csc x csc x cot x dx -1- 11 12 13 14 15 16 17 18 Derivatives of Inverse Trig Functions: d 1 [arcsin x] dx 1 x2 d 1 [arctan x] dx 1 x2 d 1 [arc sec x] dx x x2 1 d 1 [arccos x] dx 1 x2 d 1 [arc cot x] dx 1 x2 d 1 [arc csc x] dx x x2 1 Derivatives of Exponential and Logarithmic Functions: d 1 d 1 [ln x] , x 0 [log a x] dx x dx x ln a d x d x [e ] e x [a ] a x ln a dx dx Justifications for horizontal tangent lines: dy 0. f ( x ) has horizontal tangents when dx Chain Rule: dy dy du d f ( g ( x)) f '( g ( x)) g '( x) dx du dx dx Product Rule: d f ( x) g ( x) f ( x) g '( x) g ( x) f '( x) dx Quotient Rule: d f ( x) g ( x) f '( x) f ( x) g '( x) 2 dx g ( x) g ( x) Derivatives of Inverse Functions: The derivative of an inverse function is the reciprocal of the derivative of the original function at the “matching” point. 1 If (a, b) is on f ( x) , then (b, a) is on f 1 ( x) and ( f 1 ) '(b) . f '(a) Justifications for vertical tangent lines: dy f ( x ) has vertical tangents when is undefined. dx Justifications for Particle Motion: Particle is moving right/up because v(t ) 0 (positive). Particle is moving left/down because v(t ) 0 (negative). 19 Particle is speeding up (|velocity| is getting bigger) because v (t ) and a(t ) have same sign. Particle is slowing down (|velocity| is getting smaller) because v (t ) and a(t ) have different signs. AB Calculus Formulas & Justifications -2- 20 21 22 23 Intermediate Value Theorem: If f is continuous on [a, b] and k is any number between f (a ) and f (b ) , then there is at least one number c between a and b such that f (c ) k . Extreme Value Theorem: If f is continuous on the closed interval [a, b], then f has both a minimum and a maximum on the closed interval [a, b]. Justification for an Absolute Extrema. 1. Find critical numbers. 2. Identify endpoints. 3. Find f (critical numbers) and f (endpoints) . 4. Determine absolute max/min values by comparing the y-values. State in a sentence. Mean Value Theorem: If f is continuous on [a, b] and differentiable on (a, b) then there exists a f (b) f (a ) number c on (a, b) such that f '(c) . ba (Calculus slope = Algebra Slope) 24 Rolle’s Theorem: If f is continuous on [a, b] and differentiable on (a, b) and if f (a ) f (b) , 25 Justification for a Critical Number: x c is a critical number because f '( x) 0 or f '( x) is undefined. 26 then there exists a number c on (a, b) such that f '(c) 0 . Justification for Increasing/Decreasing Intervals: Inc: f ( x) is increasing on [____, ____] b/c f '( x) 0 . Dec: f ( x) is decreasing on [____, ____] b/c f '( x) 0 . 27 Justification for a Relative Max/Min Using 1st Derivative Test: Local Max: f '( x) changes from + to -. 28 Justification for Relative Max/Min Using 2nd Derivative Test: Local Max: f '(c) 0 (or und) and f ''( x) 0 . 29 Justification for a Point of Inflection: Using 2nd derivative: f ''( x) 0 (or dne) AND f ''( x) changes sign. 30 Justification for Concave Up/Concave Down: Concave Up: f ( x) is concave up on (____, ____) because f ''( x) 0 . Local Min: f '( x) changes from - to +. Local Min: f '(c) 0 (or und) and f ''( x) 0 . Using 1st derivative: f ''( x) 0 (or dne) AND slope of f '( x) changes sign. Concave Down: f ( x) is concave down on (____, ____) because f ''( x) 0 . AB Calculus Formulas & Justifications -3- 31 Justifications for linear approximation estimates: A linear approximation (tangent line) is an overestimate if the curve is concave down. A linear approximation (tangent line) is an underestimate if the curve is concave up. Integration Rules: 1 n1 x n dx x C cos xdx sin x C n 1 1 cos kx dx k sin kx C sin xdx cos x C 1 2 sin kx dx k cos kx C sec xdx tan x C 2 csc xdx cot x C sec x tan xdx sec x C 32 1 x dx ln x C e dx e C csc x cot xdx csc x C tan xdx ln cos x C e 33 34 kx dx x 1 kx e C k x 2 x 1 1 x dx arctan C 2 a a a Justifications for Reimann Sums: Left-Riemann Sums: The sum is an overestimate if the curve is decreasing. The sum is an underestimate if the curve is increasing. Right-Riemann Sums: The sum is an overestimate if the curve is increasing. The sum is an underestimate if the curve is decreasing. First Fundamental Theorem of Calculus: b a f '( x)dx f (b) f (a) (Finds the signed area between a curve and the x-axis) Properties of Integrals: 35 b f ( x) g ( x)dx a b f ( x) g ( x)dx a b cf ( x)dx c a b f ( x)dx a a b f ( x)dx a b f ( x)dx a b g ( x)dx a b g ( x)dx a b f ( x)dx a a f ( x)dx b f ( x)dx 0 a AB Calculus Formulas & Justifications -4- 36 37 38 Average Value of a Function: b 1 f avg f ( x)dx ba a Second Fundamental Theorem of Calculus: d x d g ( x) f ( t ) dt f ( x ) f (t )dt f ( g ( x)) g '( x) dx a dx a “Net Change” Theorem: b f ( x)dx represents the “net change” in the function f from time a to b. a Finding Total Amount: 39 40 41 f (b) f (a) b f '( x)dx (want = have + integral) a Steps for Solving Differential Equations: “Find a solution (or solve) the separable differentiable equation…” 1. Separate the variables 2. Integrate each side 3. Make sure to put C on side with independent variable (normally x) 4. Plug in initial condition and solve for C (if given) 5. Solve for the dependent variable (normally y) Exponential Growth and Decay: “The rate of change of a quantity is directly proportional to that quantity” dy Gives the differential equation: ky dt Which can be solved to yield: y Cekt Particle Motion Formulas: Velocity: v(t ) s '(t ) Speed: speed v(t ) Acceleration: a(t ) v '(t ) s ''(t ) s (b) s (a ) ba b 1 v(t )dt (given v (t ) ) ba a v (b) v (a ) Average Acceleration: (given v (t ) ) ba b 1 a (t )dt (given a(t ) ) ba a Average Velocity: (given s (t ) ) 42 b Displacement: Total Distance: Position at b: s (b) s (a) v(t )dt AB Calculus Formulas & Justifications a b a v(t )dt v(t ) dt b a -5- Areas in a Plane: Perpendicular to x-axis: f ( x) g ( x) dx b a f ( x ) is top curve, g ( x ) is bottom curve, a and b are x-coordinates of 43 point of intersection Perpendicular to y-axis: f ( y) g ( y) dy b a f ( y ) is right curve, g ( y ) is left curve, a and b are y-coordinates of point of intersection Steps to Finding Volume: Volume = Area 1. decide on whether it’s a dx or dy 2. find a formula for the area in terms of x or y 3. find the limits (making sure they match x or y) 4. integrate and evaluate 44 Volumes Around a Horizontal Axis of Rotation or Perpendicular to x-axis: 45 b Disc: V r 2 dx a and b are x-coordinates Washer: V R 2 r 2 dx a b a and b are x-coordinates a Slab (Cross Section): b V A( x)dx a A( x ) is the area formula for the cross section Volumes Around a Vertical Axis of Rotation or Perpendicular to y-axis: 46 b Disc: V r 2 dy a and b are y-coordinates Washer: V R 2 r 2 dy a b a and b are y-coordinates a Slab (Cross Section): b V A( y )dy a A( y ) is the area formula for the cross section Volume= AB Calculus Formulas & Justifications Surface Area= 6𝑠 2 , 2𝜋𝑟ℎ + 2𝜋𝑟 2 , 𝐷𝑜𝑛′ 𝑡 𝑛𝑒𝑒𝑑 4𝜋𝑟 2 -6-