STUFF YOU MUST KNOW COLD by AP Calculus BC Midterm

Alternative Definition of the Derivative

f’(c) = lim [f(x)-f(c)]/(x-c) as x → c

Intermediate Value Theorem

If the functions f(x) is continuous on [a,b], and y is a number between f(a) and f(b), then there exists at least one number x = c in the open interval (a,b) such that f(c) = y.

Mean Value Theorem

If the functions f(x) is continuous on [a,b], AND the first derivative exists on the interval (a,b) then there is at least one number x = c in (a,b) such that f’(c) = [f(b)-f(a)] / (b-a)

derivative of sinx

cosx

derivative of cosx

-sinx

derivative of e^x

e^x

derivative of lnx

1/x

derivative of log_ax

1/[x*(lna)]

derivative of tanx

sec²x

derivative of cotx

-csc²x

derivative of secx

secxtanx

derivative of cscx

-cscxcotx

derivative of sinu

(cosu)u’

derivative of tanu

(sec²u)u’

derivative of secu

(secu*tanu)u’

derivative of cosu

-(sinu)u’

derivative of tanu

(sec²u)u’

derivative of cotu

-(csc²u)u’

derivative of secu

(secu*tanu)u’

derivative of cscu

-(cscu*cotu)u’

derivative of lnu

1/u * u’

derivative of e^u

e^u * u’

derivative of a^x

(lna)a^x

derivative of a^u

(lna)a^uu’

derivative of log_au

1/(lna)u * u’

derivative of arcsin(u/a) or sin^-1(u/a)

1/sqrt(a²-u²)

derivative of arccos(u/a) or cos^-1(u/a)

-1/sqrt(a²-u²)

derivative of arctan(u/a) or tan^-1(u/a)

1/(a²+u²)

derivative fo arccot(u/a) or cot^-1(u/a)

-1/(a²+u²)

derivative of arcsecx or sec^-1(u/a)

1/(lul*sqrt(u²-a²))

derivative of arccsc(u/a) or csc^—1(u/a)

-1/(lul*sqrt(u²-a²))

derivative of an inverse function

[f^-1(y)]' = 1/(f’(x)) if f(x) = y

Rolle’s Theorem

If the function f(x) is continuous on [a,b], AND the first derivative exists on the interval (a,b) AND f(a) = f(b), then there is at least one number x = c in (a,b) such that f’(c) = 0.

Extreme Value Theorem

If the function f(x) is continuous on [a,b], then the functions is guaranteed to have an absolute maximum and an absolute minimum on the interval.

(Derivative of an Inverse Function) If f has an inverse function g then:

g’(x) = 1/ f’(g(x)) ; derivatives are reciprocal slopes

Implicit Differentiation

replace dy/dx for each y; isolate dy/dx; when taking second derivative (d²y/d²x), substitute dy/dx in when needed

Average Rate of Change (AROC)

m_sec = [f(b)-f(a)] / (b-a)

Instantaneous Rate of Change (IROC)

m_tan = f’(x) = lim[ f(x+h) - f(x)] / h as h → 0

Critical Point

dy/dx or f’(x) = 0 or undefined

Local Minimum

dy/dx or f’(x) goes from (-, 0, +) or (-, und, +) OR d²y/dx² > 0

Local Maximum

dy/dx or f’(x) goes (+, 0, -) or (+, und, -) OR d²y/dx² < 0

Points of Inflection (POI)

concavity changes; d²y/dx² or f’’(x) goes from (+, 0, -), (-,0,+), (+, und, -), OR (-, und, +)

f’(x) > 0

f(x) is increasing

f’(x) < 0

f(x) is decreasing

Relative Maximum

f’(x) = 0 or DNE and sign f’(x) changes from + to -

Relative Minimum

f’(x) = 0 or DNA and sign of f’(x) changes from - to +

Absolute Maximum or Minimum

follow relative min/max steps with critical points and endpoints; plug x value into original equation to find y-value since max value is a Y-VALUE

f’’(x) > 0

f(x) is concave up

f’’(x) < 0

f(x) is concave down

f’(x) = 0 and sign of f’’(x) changes

POI at x

relative maximum with f’’(x)

f’’(x) < 0

relative minimum with f’’(x)

f’’(x) > 0

Equation of a tangent line at a point

y₂ - y₁ = m_tan (x₂ - x₁); need a slope (derivative) and point

If the largest exponent in the numerator < largest exponent in the denominator, then limf(x) as x → ± ∞ =

0

If the largest exponent in the numerator > largest exponent in the denominator, then limf(x) as x → ± ∞ =

DNE

If the largest exponent in the numerator = largest exponent in the denominator, then limf(x) as x → ± ∞ =

a/b

What four things can you do no a calculator that needs no work shown?

graphing a function within an arbitrary view window; finding zeros of a function; computing the derivative of a function numerically; computing the definite integral of a function numerically

s(t)

position function

v(t)

velocity function

a(t)

acceleration function

s’(t) =

v(t)

s’’(t)

a(t) = v’(t)

∫a(t)

v(t) = s’(t)

∫ v(t)

s(t)

speed =

l v(t) l

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

speed is increasing

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

speed is decreasing

v(t) > 0

moving right

v(t) < 0

moving left

Displacement

∫ v(t) dt from [t₀, t_f]

total distance

∫ l v(t) l dt from [t₀, t_f]

What is the unit of position, velocity, and acceleration?

ft, ft/sec, ft/sec²

Average Velocity

(final position - initial position) / (total time) = ∆x/∆t

Accumulation

x(0) +∫ v(t)dt from [t₀, t_f]

ln N = p so N =

e^p

ln e =

1

ln 1 =

0

ln(MN) =

lnM + lnN

ln(M/N) =

lnM - lnN

p * lnM =

ln(M^p)

What equation do you use with y is a differentiable function of t such that y > 0 and y’ = ky and the rate of change of y is proportional to y?

y = Ce^kt

When solving differential equation, what do you do?

separate variables; integrate; add +C to one side; use initial conditions to find “C”; write equation in form of y = f(x)

What do you always add when integrating an equation with infinite endpoints?

+ C

Fundamental Theorem of Calculus

∫_a^b f(x) dx = F(b) - F(a) where F’(x) = f(x)

d/dx ∫_a^u f(t) dt =

f(u) du/dx

d/dx ∫ f(t) dt as h(x) → g(x) =

f(g(x))g’(x) - f(h(x))h’(x)

Riemann Sums

means rectangular approximation; DO NOT EVALUATE THE INTEGRAL, just add up the areas of the rectangles

Trapezoidal Rule

A_trap = ([b₁ + b₂]/2) * h

sin²x + cos²x =

1

1 + tan²x =

sec²x

cot²x + 1 =

csc²x

sin2x =

2 sinxcosx

cos2x =

cos²x - sin²x or 1-2sin²x

cos²x =

½ * (1 + cos2x)

sin²x =

½ * (1 - cos2x)

tan x =

sinx/cosx

cotx =

cosx/sinx

cscx =

1/sinx

secx =

1/cosx

∫ du =

u + C