AP Calculus Flashcards

AP Calculus Flashcards

memorize.aimemorize.ai (lvl 286)
Section 1

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y= constant

Front

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Cards (77)

Section 1

(50 cards)

y= constant

Front

y' = 0

Back

f(x)= function f'(x)= inverse function f(f⁻¹(x)) = x

Front

f'(f⁻¹(x))[f⁻¹(x)]' = 1 [f⁻¹(x)]' = 1/ f'(f⁻¹(x))

Back

Speed

Front

| u(t) |

Back

y= e^x

Front

dy/dx= e^x × d(x)/dx

Back

y= cosx

Front

dy/dx= -sinx × d(x)/dx

Back

y= arccotu

Front

Back

Relationship between Differentiability and Continuity

Front

Differentiability implies continuity

Back

Extreme Value Theorem

Front

If f is continuous on a closed interval [a,b] then f has both a minimum and a maximum on the interval

Back

Instantaneous Rate of Change Slope of Tangent Line

Front

Lim ∆x→0 f(x+∆x)-f(x)/∆x Lim h→0 f(x+h)-f(h)/h

Back

Trig Limit (sinx)

Front

Squeeze Thm Lim θ→0 sinθ/θ = 1

Back

y= arctanu

Front

Back

y= arcsecu

Front

Back

Normal Line

Front

Normal line is a line parallel to the tangent line at the point of tangency.

Back

General Position Formula for Free-Falling Objects

Front

x(t)= (1/2)gt²+Vt₀+S₀ Meters: g = -9.8m/sec² Feet: g = -32ft/sec² V₀= initial velocity S₀= initial height g= gravity

Back

Maximum height of an object

Front

1) v(t)= 0 Solve for t 2) Plug that t into position function

Back

y= cotx

Front

y'=-csc^2(x)

Back

Chain Rule y= f(g(x))

Front

y' = f'(g(x)) × g'(x)

Back

3 reasons limits don't exist

Front

1. They are unbounded: Lim x→0 1/x² 2. Right-hand limit ≠ Left-hand limit: Lim x→0 |x|/x 3. Oscillating Behavior: Lim x→0 sin(1/x)

Back

Quotient Rule

Front

y = f(x)/g(x) y'= g(x)f'(x) - f(x)g'(x)/ [g(x)]²

Back

Power rule y=x^n

Front

dy/dx = nxⁿ⁻¹ × d(x)/dx

Back

Technique for Evaluating Limits

Front

1) Direct Substitution 2) Factoring 3) Long/Synthetic Division 4) Expand 5) Common Denominators 6) Multiply by the Conjugate

Back

y= cscx

Front

y'=-cscxcotx

Back

y= lnu

Front

y'= 1/u • u'

Back

Vertical Tangent

Front

dy/dx doesn't exist

Back

Acceleration

Front

Derivative of velocity

Back

3 reasons derivatives do not exist

Front

1. Sharp Turn 2. Vertical Tangent 3. Discontinuity

Back

Product Rule

Front

y = f(x)g(x) y' = f(x)g'(x) + f'(x)g(x)

Back

Average Rate of Change Slope of Secant

Front

= Y₁-Y₂/X₁-X₂ f(x+∆x)-f(x)/∆x

Back

Horizontal Tangent

Front

dy/dx= 0

Back

Speed Decreasing

Front

r(t) and a(t) have different signs

Back

Intermediate Value Theorem

Front

If f is continuous on the closed interval [a,b]. f(a)≠f(b), and k is any number between f(a) and f(b), then there is at least one number c in [a,b] such that f(c)=k

Back

y=sinx

Front

dy/dx= cosx × d(x)/dx

Back

Continuity at a Point

Front

Given that f(x) at x=a I. f(a) must exist II. Lim f(x) x→a exists III. Lim f(x) x→a = f(a)

Back

Speed Increasing

Front

v(t) and a(t) have the same sign

Back

Critical Number

Front

Let f be defined at c. If f'(c)= 0 or if f is not differentiable at c, then c is a critical number

Back

a) particle at rest b) particle to right/up c) particle to left/down

Front

a) v(t)= 0 b) v(t) > 0 c) v(t) < 0

Back

vertical asymptote

Front

Non-removable discontinuities

Back

y= tanx

Front

y'= sec^2(x)

Back

Instantaneous Velocity (Velocity= Speed and Direction)

Front

Derivative of the position function

Back

y= a^u (Constant to a Variable Power)

Front

y= lna × a^u × u'

Back

Hits the ground

Front

1) x(t)= 0 2) Plug that t into position function

Back

y= arccosu

Front

Back

Average Function

Front

The change in position/The change in time

Back

Trig Limit (cosx)

Front

Lim θ→0 1-cosθ/θ = 0

Back

y= secx

Front

y'= secxtanx

Back

Unit Circle

Front

Back

y= arccscu

Front

Back

Trig Table

Front

Back

y= log∨a(u)

Front

y'= (1/lna × U) × U'

Back

y= arcsinu

Front

Back

Section 2

(27 cards)

The graph of f(x) is concave up

Front

f''(x) > 0

Back

∫ 1/x dx

Front

ln |x| +c

Back

Fundamental Thm of Calculus

Front

Back

∫ sin du

Front

-cos u +c

Back

Guidelines for Finding Extreme on a Closed Interval

Front

1) Find the critical numbers of f in (a,b) 2) Evaluate f at each critical number of [a,b] 3) Evaluate f at each endpoint of [a,b] 4) The least of these values is the minimum and the greatest is the maximum

Back

Graph of f(x) is decreasing

Front

f'(x) < 0

Back

∫ a^u du

Front

Back

Point of inflection

Front

The graph of f(x) changes concavity, f''(x)=0 or undefined, f''(x) changes sign

Back

∫ dx

Front

x+c

Back

The graph of f(x) is concave down

Front

f''(x) < 0

Back

∫ cos u du

Front

sinu +c

Back

Horizontal Asymptotes

Front

y= lim f(x) x → ± ∞

Back

Second Fundamental Theorem of Calculus

Front

Back

∫ x^n dx

Front

x^(n+1)/(n+1) + c; n≠-1

Back

Average Value of a Function

Front

Back

∫ secutanu du

Front

sec u +c

Back

Max

Front

f'(x) goes from + to -

Back

Rolle's Thm

Front

Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number c in (a,b) such that f'(c)=0

Back

Graph of f(x) is increasing

Front

f'(x) > 0

Back

Second Derivative Test

Front

1) Find critical #s from First Derivative 2) Take f'(x) 3) Plug points into f''(x) 4) f''(x) > 0 Min pt f''(x) < 0 Max pt f''(x) = 0 Test fails

Back

∫ csc^2 u du

Front

-cot u + c

Back

∫ cscucotu du

Front

-csc u+ c

Back

Mean Value Thm

Front

If f is continuous on [a,b] and differentiable on the open interval (a,b) then there exists a number c such that f'(c)= f(b)-f(a)/b-a

Back

∫ e^u du

Front

e^u +c

Back

Min

Front

f'(x) goes from - to +

Back

∫ sec^2 u du

Front

tan u+ c

Back

Trig Identities

Front

1). sin^2x+ cos^2x= 1 a) ÷sin^2x b) ÷cos^2x 2) sin2x= 2sinxcosx 3) cos2x= cos^2x-sin^2x OR = 1-2sin^2x OR = 2cos^2x-1

Back