AP BC Calculus: Proficiency Formulas (S1)

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Intermediate Value Theorem

If f(1)=-4 and f(6)=9, then there must be a x-value between 1 and 6 where f crosses the x-axis.

Average Rate of Change

Slope of secant line between two points, use to estimate instantanous rate of change at a point.

Instantenous Rate of Change

Slope of tangent line at a point, value of derivative at a point

Formal definition of derivative

limit as h approaches 0 of [f(a+h)-f(a)]/h

Alternate definition of derivative

limit as x approaches a of [f(x)-f(a)]/(x-a)

When f '(x) is positive, f(x) is

increasing

When f '(x) is negative, f(x) is

decreasing

When f '(x) changes from negative to positive, f(x) has a

relative minimum

When f '(x) changes fro positive to negative, f(x) has a

relative maximum

When f '(x) is increasing, f(x) is

concave up

When f '(x) is decreasing, f(x) is

concave down

When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x) has a

point of inflection

When is a function not differentiable

corner, cusp, vertical tangent, discontinuity

Product Rule

uv' + vu'

Quotient Rule

(uv'-vu')/v²

Chain Rule

f '(g(x)) g'(x)

Particle is moving to the right/up

velocity is positive

Particle is moving to the left/down

velocity is negative

absolute value of velocity

speed

y = sin(x), y' =

y' = cos(x)

y = cos(x), y' =

y' = -sin(x)

y = tan(x), y' =

y' = sec²(x)

y = csc(x), y' =

y' = -csc(x)cot(x)

y = sec(x), y' =

y' = sec(x)tan(x)

y = cot(x), y' =

y' = -csc²(x)

y = sin⁻¹(x), y' =

y' = 1/√(1 - x²)

y = cos⁻¹(x), y' =

y' = -1/√(1 - x²)

y = tan⁻¹(x), y' =

y' = 1/(1 + x²)

y = cot⁻¹(x), y' =

y' = -1/(1 + x²)

y = e^x, y' =

y' = e^x

y = a^x, y' =

y' = a^x ln(a)

y = ln(x), y' =

y' = 1/x

y = log (base a) x, y' =

y' = 1/(x lna)

To find absolute maximum on closed interval [a, b], you must consider...

critical points and endpoints

mean value theorem

if f(x) is continuous and differentiable, slope of tangent line equals slope of secant line at least once in the interval (a, b)

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

If f '(x) = 0 and f"(x) > 0,

f(x) has a relative minimum

If f '(x) = 0 and f"(x) < 0,

f(x) has a relative maximum

Linearization

use tangent line to approximate values of the function

left riemann sum

use rectangles with left-endpoints to evaluate integral (estimate area)

right riemann sum

use rectangles with right-endpoints to evaluate integrals (estimate area)

trapezoidal rule

use trapezoids to evaluate integrals (estimate area)

average value of f(x)

= 1/(b-a) ∫ f(x) dx on interval a to b

If g(x) = ∫ f(t) dt on interval 2 to x, then g'(x) =

g'(x) = f(x)

Fundamental Theorem of Calculus

∫ f(x) dx on interval a to b = F(b) - F(a)

To find particular solution to differential equation, dy/dx = x/y

separate variables, integrate + C, use initial condition to find C, solve for y

To draw a slope field,

plug (x,y) coordinates into differential equation, draw short segments representing slope at each point

methods of integration

substitution, parts, partial fractions

∫ u dv =

uv - ∫ v du

dP/dt = kP(M - P)

logistic differential equation, M = carrying capacity

P = M / (1 + Ae^(-Mkt))

logistic growth equation

volume of solid with base in the plane and given cross-section

∫ A(x) dx over interval a to b, where A(x) is the area of the given cross-section in terms of x

volume of solid of revolution - no washer

π ∫ r² dx over interval a to b, where r = distance from curve to axis of revolution

volume of solid of revolution - washer

π ∫ R² - r² dx over interval a to b, where R = distance from outside curve to axis of revolution, r = distance from inside curve to axis of revolution

length of curve

∫ √(1 + (dy/dx)²) dx over interval a to b

L'Hopitals rule

use to find indeterminate limits, find derivative of numerator and denominator separately then evaluate limit

indeterminate forms

0/0, ∞/∞, ∞*0, ∞ - ∞, 1^∞, 0⁰, ∞⁰

derivative of parametrically defined curve
x(t) and y(t)

dy/dx = dy/dt / dx/dt

second derivative of parametrically defined curve

find first derivative, dy/dx = dy/dt / dx/dt, then find derivative of first derivative, then divide by dx/dt

length of parametric curve

∫ √ (dx/dt)² + (dy/dt)² over interval from a to b

given velocity vectors dx/dt and dy/dt, find speed

√(dx/dt)² + (dy/dt)² not an integral!

given velocity vectors dx/dt and dy/dt, find total distance travelled

∫ √ (dx/dt)² + (dy/dt)² over interval from a to b

area inside polar curve

1/2 ∫ r² over interval from a to b, find a & b by setting r = 0, solve for theta

area inside one polar curve and outside another polar curve

1/2 ∫ R² - r² over interval from a to b, find a & b by setting equations equal, solve for theta.

Definition of Continuity

A function is continuous if 1) f(c) is defined. 2) lim f(x) as x approaches c exists 3) lim f(x) as x approached c equals f(c)

d/dx(arcsecx)=

y'= x'/(I x I*√(x² - 1))

Fundamental Theorem of Calculus Part 2

if f(x) is continuous on an open interval I, then d/dx[∫f(t)dt] = f(x)

∫sinxdx

-cosx+ C

∫tanxdx

-ln IcosxI + C

∫secxdx

ln Isecx + tanxI + C

∫secxtanxdx

secx + C

∫csc^2xdx

-cotx + C

∫a^xdx

(1/lna)(a^x) + C

∫1/(a^2 + x^2)dx

(1/a)arctanx/a + C

∫cosxdx

sinx + C

∫cotxdx

ln IsinxI + C

∫cscxdx

-ln Icscx + cotxI + C

∫sec^2xdx

tanx + C

∫cscxcotxdx

-cscx + C

∫1/ (a^2 - x^2)dx

arcsin (x/a) + C

sin^2x=

(1-cos2x)/2

cos^2x=

(1+cos2x)/2

Parametric Form of 2nd Derivative

(d/dt(dy/dx))/(dx/dt)

Rolle's Theorem

If f(x) is continuous on the closed interval [a, b], differentiable on (a, b), and satisfies f(a) = f(b), then for some c in the interval (a, b), we have f'(c) = 0

Second Derivative Test

if f'(c) = 0 and f''(c) > 0 then minimum; if f'(c) = 0 and f''(c) < 0 then maximum

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