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AP Calculus Exam

Normal Line line perpendicular to the tangent line at the point of tangency
Derivative Of Sine cosine
Derivative Of Cosine -sine
Derivative Of Tangent sec^2
Derivative Of Cosecant -csc cot
Derivative Of Secant sec tan
Derivative Of Cotangent -csc^2
When F'(x)=0, F(x)... has a CV
When F(x) Has A Horizontal Tangent, F'(x)... is equal to 0
Equation Of A Tangent Line y2-y1=slope(x2-x1)
Product Rule [f'(x) ⋅ g(x)] + [f(x) ⋅ g'(x)]
Slope Formula (y2-y1) / (x2-x1)
Quotient Rule { [f'(x) ⋅ g(x)] + [f(x) ⋅ g'(x)] } / [g(x)^2]
Chain Rule f'[g(x)] ⋅ g'(x) = derive outside, leave inside, derive inside
Derivative Of e^any e^any ⋅ derivative(any)
Derivative Of ln(any) 1/any ⋅ derivative(any)
Derivative Of a^(any) a^any ⋅ ln(any) ⋅ derivative(any)
Derivative Of log_a_(any) 1/any ⋅ 1/lna ⋅ derivative(any)
In Order For A Function To Be Differentiable... the function must be continuous & must have no cusps
When F''(x)=0 or DNE, F(x)... has a POSSIBLE point of inflection
When F''(x)=0 or DNE, F'(x)... has a CV
When F''(x)>0, F(x)... is concave up
When F''(x) Goes From Positive To Negative, F(x)... has a POI
When F''(x)<0, F(x)... is concave down
When F''(x) Goes From Negative To Positive, F(x)... has a POI
When F''(x) Goes From Positive To Negative, F'(x)... has a relative max
When F''(x) Goes From Negative To Positive, F'(x)... has a relative min
Extreme Value Theorem if f(x) is continuous on [a,b], than f(x) is guaranteed to have an absolute maximum & an absolute minimum on [a,b]
Absolute Extrema Can Occur... at endpoints & at CVs
Relative Extrema Can Occur... only at CVs
To Find The Absolute Extrema From An Analytical Function... check if it's continuous, & then use a candidates test
Average Rate Of Change slope of secant --> (y2-y1)/(x2-x1)
Instantaneous Rate Of Change slope of tangent line --> f'(x)
Mean Value Theorem if f(x) is continuous & differentiable on [a.b], then it is guaranteed to exist a value of c where f'(c)= [f(b)-f(a)] / [b-a]
Intermediate Value Theorem if f(x) is continuous on [a,b], & f(a) ≤ k ≤ f(b), then there exists at least one value, x=c, on [a,b] such that=at f(c)=k
When You Derive "y"... put a y' behind it
Steps Of Deriving 1) multiply 2) subtract one from the exponent
Steps Of Integration 1) add one to exponent 2) divide by new exponent
Integral Of Cosine sinx
Integral Of Sine -cosx
Steps Of U-Substitution 1) set "u" equal to inside function # 2) find the derivative of "u" # 3) move coefficients to the "du" side # 4) plug "u" & "du" into the integral # 5) plug original boundaries into "u" to get "u" boundaries # 6) solve # 7) plug original function back in
When Left Hand Riemann Sum Is Increasing... it is an under approximation
When Left Hand Riemann Sum Is Decreasing... it is an over approximation
When Right Hand Riemann Sum Is Increasing... it is an over approximation
When Right Hand Riemann Sum Is Decreasing... it is an under approximation
Trapezoidal Riemann Sums (left hand + right hand) / 2
Second Fundamental Theorem Of Calculus take the "x" boundary, plug it in for t, and multiply it by the derivative of the "x" boundary
Average Velocity Equation p(b)-p(a) / b-a = slope of position OR (1/b-a) ∫a(t)dt
Average Acceleration Equation v(b)-v(a) / b-a = slope of velocity
When Velocity Is>0, Position... is x to the right & y up
When Velocity Is<0, Position... is to the left & down
When Velocity Goes From Positive To Negative Or Vice Versa, Position... changes direction
When Acceleration>0, Velocity... is increasing
When Acceleration<0, Velocity... is decreasing
When Acceleration Changes From Positive To Negative, Velocity... slows down
When Acceleration Changes From Negative To Positive, Velocity... speeds up
Speed Is The Absolute Value Of... velocity
Velocity & Acceleration In Relation To Speed same signs = increasing speed // different signs = decreasing speed
Net Distance Equation ∫v(t)dt
Total Distance Equation ∫|v(t)|dt
Area Equation ∫(TOP-BOTTOM)dx
Volume Equation π∫(OUTER-AXIS)^2-(INNER-AXIS)^2dx
Cross Section Volume value ⋅ ∫(TOP-BOTTOM)^2
Cross Section Value Of A Square 1
Cross Section Value Of Semi-Circle π/8
Cross Section Value Of Isosceles Triangle 1/2
Cross Section Value Of Equilateral Triangle √3/4
Area In Respect To "y" ∫(RIGHT-LEFT)dy & use y-limits
Steps For Solving Differential Equations 1) separate the variables -- 2) integrate both sides -- 3) find c -- 4) use c to solve for y
Created by: abievans
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