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AP Calculus Semester 1 Review

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Section 1

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y' (csc x)

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Calculus Math

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Mar 1, 2020

Cards (48)

Section 1

(48 cards)

y' (csc x)

Front

-csc xcot x

Back

sin (2θ) =

Front

2sinθcosθ

Back

mean value theorem

Front

if y=f(x) is continuous at every point of the closed interval [a,b] and differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which the instantaneous rate of change equals the mean rate of change

Back

derivative test of marginal cost

Front

P' = R' -C'

Back

cos² (x) =

Front

0.5 + 0.5cos(2x)

Back

L(x) =

Front

f(a) + f'(a)(x-a)

Back

instantaneous rate of change

Front

limit of the difference quotient; slope of the tangent line = limh→0 f(h+x) - f(x)/h (derivative)

Back

y'(cos^-1(x))

Front

-1/√(1-x^2)

Back

composition of functions [chain rule]

Front

(fog)(x)

Back

speed =

Front

|velocity| = |s'(t)|

Back

product rule

Front

d/dx (p•q) = p' • q + p • q'

Back

d/dx (c) =

Front

0

Back

AP way to identify absolute max and min values on a given interval

Front

use a chart to find the y values of the critical points and end point of the given interval

Back

y'(a^x)

Front

a^x • ln(a)

Back

y'(tan^-1(x))

Front

1/(1+x^2)

Back

y' (sin x)

Front

cos x

Back

y' (cot x)

Front

-csc^2 x

Back

y'(loga x)

Front

1/(x • ln(a))

Back

y'(e^x)

Front

e^x

Back

y' (tan x)

Front

sec^2 x

Back

y'(csc^-1(x))

Front

-1/|x|√(x^2-1)

Back

Rolle's Theorem

Front

let "f" be continuous on a closed interval [a,b] and differentiable on (a,b) with f(a)=f(b); there is at least one point "c" in (a,b) such that f'(c)=0

Back

for R(x) = (ax^T + ...)/(bx^B + ...), then R(x) has a horizontal asymptote at the line...

Front

1: y=0 if B>T; 2: y=a/b if T=B; 3: y=mx+b if T=B+1; 4: none if T>B+1

Back

4 types of differentiability

Front

a function will fail to be differentiable at a point where there is a corner, cusp, vertical tangent, or discontinutiy

Back

Vav = displacement/travel time =

Front

Δs/Δt = f(t+Δt) - f(t)/Δt

Back

sin² (x) =

Front

0.5 - 0.5cos(2x)

Back

4 types of discontinuity

Front

removable, jump, infinite, and oscillating

Back

d/dx (f±g) =

Front

f'±g'

Back

extreme value theorem

Front

if f is continuous on a closed interval [a,b], then f had both a maximum value and a minimum value on the interval

Back

sin² θ + cos² θ =

Front

1

Back

quotient rule

Front

d/dx (high/low) = low(high)' - high(low)'/(low)²

Back

mean rate of change

Front

f'(c) = f(b)-f(a)/(b-a)

Back

y'(sec^-1(x))

Front

1/|x|√(x^2-1)

Back

Intermediate Value Theorem

Front

IF 1. f is a continuous function on the closed interval [a,b] AND 2. k is any y-value between f(a) and f(b), that is f(a)<k<f(b), THEN 3. there is at least one number c in [a,b] that is a≤c≤b, so that f(c)=k

Back

y'(ln x)

Front

1/x

Back

y' (cos x)

Front

-sin x

Back

s =

Front

f(t)

Back

y'(cot^-1(x))

Front

-1/(1+x^2)

Back

cos (2θ) = cos² θ - sin² θ =

Front

2cos² θ - 1 = 1 - 2sin² θ

Back

derivative

Front

the derivative of the function f with respect to the variable x is the function f' whose value at x is f'(x) = limh→0 f(x+h) - f(x)/h, provided the limit exists OR the derivative of the function f at the point x=a is f'(a) = limx→a f(x)-f(a)/(x-a), provided the limit exists

Back

y' (sec x)

Front

sec xtan x

Back

if y=f(u) and u=g(x), [chain rule]

Front

then dy/dx = (dy/du) • (du/dx)

Back

d/dx (x^n) =

Front

nx^(n-1)

Back

d/dx [(fog)(x)] = [chain rule]

Front

f'(g(x)) • g'(x)

Back

average rate of change

Front

difference quotient; slope of secant line = f(h+x) - f(x)/h

Back

AP Way to Show Continuity

Front

1. Point exists in Domain (find f(c)) 2. Prove limit exists (limx→c^+ = limx→c^-) 3. Show limx→c = f(c)

Back

y'(sin^-1(x))

Front

1/√(1-x^2)

Back

d/dx (cu) =

Front

c•u'

Back