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AP Calculus Exam

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