AP calc ab unit 1 & 2 formulas/rules

sum and difference limit rule

lim_x→c (f(x) +- g(x)) = lim_x→c f(x) +- lim_x→c g(x)

product limit rule

lim_x→c (f(x) * g(x)) = lim_x→c f(x) * lim_x→c g(x)

quotient limit rule

lim_x→c (f(x)/g(x)) = (lim_x→c f(x))/(lim_x→c g(x))

constant multiple limit rule

lim_x→c (k(f(x)) = k * lim_x→c f(x)

compositions limit rule

lim_x→c f(g(x)) = f(lim_x→cg(x))

lim_x→0 (sinx)/0

1

lim_x→0 (cosx-1)/x

0

lim_x→0 x/(sinx)

1

lim_x→0 (1-cosx)/x

0

lim_x→0 (sinx)/x

1

lim_{x→+-infinity} (sinx)/x

0

lim_{x→+-infinity} (cosx)/x

0

horizontal asymptote rules (m = degree of num, n = degree of den)

if m>n, no HA

if m=n, HA at y = (leading coeff of m)/(leading coeff of n)

if m<n, HA at y=0

infinite limit (VAs)

if lim_x→c^- f(x) = +-infinity OR if lim_x→c⁺ f(x) = +-infinity, there is a VA at x=c

limits at infinity (HAs)

if lim_x→infinity f(x) = c OR if lim_{x→-infinity} f(x) = c, there is a HA at y=c

definition of continuity

a function f(x) is continuous at x=c if & only if

1. f(c) exists,

2. lim_x→c f(x) exists

3. f(c) = lim_x→c f(x) exists

intermediate value theorem

if function f(x) is continuous on a closed interval [a,b], then if k is between f(a) and f(b), then f(c) = k for some c in [a,b]

limit definition of a derivative at point x = instantaneous rate of change = slope of tangent line

f’(x) = lim_h→0 (f(x+h) - f(x))/h

limit definition of a derivative at point x=a or instantaneous rate of change at x=a

f’(a) = lim_x→a (f(x) - f(a))/(x-a)

product rule

d/dx (f(x) g(x)) = f(x) * g’(x) + g(x) * f’(x)

quotient rule

d/dx (f(x)/g(x)) = (g(x) * f’(x) - f(x) * g’(x))/(g(x))²

average rate of change over interval [a,b] (slope of secant line)

(f(b) - f(a))/(b-a)

position

s = f(t)

velocity

ds/dt = f’(t) = v(t)

acceleration

d²s/dt² = f’’(t) = v’(t) = a(t)

speed

| v(t) |

d/dx (sinx)

cosx

d/dx (cosx)

-sinx

d/dx (tanx)

sec²x

d/dx (cotx)

-csc²x

d/dx (secx)

secx * tanx

d/dx (cscx)

-cscx * cotx

chain rule

if y=f(u) is a function of u and u=g(x) is a function of x, then d/dx (f(g(x)) = f’(g(x)) * g’(x)

formulas for deriving inverse functions

• g’(x) = 1/(f'(g(x))

• (f^-1)’(a) = 1/(f’(f^-1(a))

y=arcsin x iff

sin y = x

y = arccos x iff

cos y = x

y = arctan x iff

tan y = x

y = arccot x iff

cot y = x

y = arcsec x iff

sec y = x

y = arccsc x iff

csc y = x

d/dx (arcsin u) =

u’/sqrt(1-u²)

d/dx (arccos u) =

-u’/sqrt(1-u²)

d/dx (arctan u) =

u’/(1+u²)

d/dx (arccot u) =

-u’/(1+u²)

d/dx arcsec u =

u’(|u| sqrt(u²-1))

d/dx arccsc u =

-u’/(|u| sqrt(u²-1))

d/dx e^x

e^x

d/dx e^u

e^u * u’

d/dx a^x

a^x * ln a

d/dx a^u

a^u * ln a * u’

d/dx ln x

1/x

d/dx ln u

1/u * u’ = u’/u

d/dx log_ax

1/(x * ln a)

d/dx log_au

u’/(u * ln a)