Post
by **K** » Mon Mar 28, 2011 4:36 am

Derivatives

The derivative of a function f(x), written f^{1}(x), is a function which gives the slope to the tangent line of f(x)'s graph at point P.

Q is a point on f(x). The line that contains P and Q is a secant line.

lim (Q -> P) = f^{1}(x)

f^1(x) = lim (^x -> 0) f(x + ^x) - f(x) / ^

Position: s(t) = -16t^{2} + (initial velocity) * t + (initial position)

Replace 16 with (- 4.9) for meters.

inst accel: dv / dt = s'' = d^{2}s / dt^{2}

Product Rule: (f * g)^{1} = f^{1} * g + g^{1} * f

Quotient Rule: (f/g)^{1} = (f^{1} * g - g^{1} * f) / g^{2}

Constant Rule: (cf)^{1} = c * f^{1}

Derivatives of Trig:

sin(x) = cos(x)

cos(x) = - sin(x)

tan(x) = sec^{2}

sec(x) = sec(x)tan(x)

cot(x) = - cse^{2} (x)

cse(x) = cse(x)cot(x)

Chain Rule: (dy / dx) = (dy / du) * (du / dx)

Power Rule: d / dx (u^m) = mu^(m-1) * (du / dx)

Critical Number (or Critical Point) x = c is where f(x) potentially has a relative maximum or minimum, which is where

1. f^{1}(c) = 0

OR

2. f^{1}(c) is undefined, but f(c) is defined (steep point)

Absolute Maximum or Minimum:

Find critical numbers, solve for f(each critical number), throw endpoints into the pot, and take highest and lowest points, respectively.

Mean Value Theorum: If f is continuous and differentiable (no sharp points) on the interval [a, b], there is a point "c" between "a" and "b" where the slope of the tangent line = slope of the secant line, the latter of which joins the endpoints.

First Derivative Test: If x = c is a critical number of f and f^{1} is positive before 0 and negative after 0, f has a relative maximum at x = c. If it's a negative, then positive, it's a relative minimum. If both the sides are the same sign, f has a reflection point (or saddle point) at x = c.

Concavity: A function is increasing if its derivative is positive. If a function is increasing, it's concave upward. If f^{||} > 0, it's concave upward. If it's negative, it's concave downward. Point of inflection = point of concavity-flippage; occurs where f^{||} = 0

Second Derivative Test: Suppose that x = c is a critical number of f, i.e., f^{1}(c) = 0. If f^{||} > 0, f has a relative minimum at x = c. If the second derivative is negative, it has a maximum at x = c. If the second derivative = 0, the test fails.

Maximum/Minimum Problems: equation for subject to be max/min-ified, equation for object with given condition, algebra to one term for second equation (simplify as necessary), derivative, set derivative equal to 0 (must obtain critical point, which becomes the new value for the previous equation, solve for all values

Differentiable: For a small change in x, called delta x, the change in f, called delta f, is approximately given by (delta f = f^{1} * delta x). The smaller that delta x, the better the approximation. The percentage error (or relative error) is (delta f) / f, and we can replace delta f with f^{1} delta x.

Antidirivative = indefinite integral. Antidifferentiation = integration.

average value of f(x) over [a, b] = 1 / (b - a) of the definite interval of [a, b] of f(x) dx (same formula times 1 / (b – a))

The definite integral of f (x) over the interval of [a, b] = F(b) - F(a), where F is an antiderivative of f.

Areas are definite integrals.

SHORTCUT: Fundamental Theorem of Calculus-

lim as n approaches infinity of Sigma (above pattern) f(x_{i}^{*}) delta x = definite integral (weird S) of a and b of f(x) with respect to dx

2nd Fundamental Theorem of Calculus: If F(x) = interval of [3, x] f(t) dt,

F^{1} (x) = f(x)

HERPDERPDERP!

u - substitution (a technique for integrating):

1) u = parenthetical value

2) find du / dx (derivative = du / dx)

3) solve for dx

4) convert the integral to u terms (substitute the u for the parenthetical value and substitute the dx from the previous step)

5) put it back into x terms (substitute the u back into the designated parenthetical value)

Definate Integral Properties:

aSb = +

bSa = -

aSa = 0

bSa = ca bc

constant rule

integral of sin kx = - (1/k) cos kx

Integrating Arc Trig

S (du) / (a^{2} – u^{2})^(1/2) = sin^{1} u/a + C

S (du) / (a^{2} + u^{2}) = 1/a tan^{1} u/a + C

S (du) / u (u^{2} – a^{2})^{(1/2)} = 1/a sec^{1} |u|/a + C

dy = y^{1} dx

Hyperbolic Functions:

sinh x = e^{x} – e^{-x}

_________2

cosh x = e^{x} + e^{-x}

_________2

tanh x = e^{x} – e^{-x}

_______e^{x} + e^{-x}

cosh^{2} t – sinh^{2} t = 1

Catenary Curve:

y = a cosh (x/a)

Finding Areas Between Curves:

_{A}S^{B} [f(x) - g(x)] dx (horizontal curves)

____________" dy (vertical curves)

f(x) = function of the left/top curve, while g(x) = function of the right/bottom curve

polycurves: break it up

Numerical Integration:

b-a = interval size

n

Trapezoidal Rule: _{A}S^{B} f(x) dx is the rough equivalent of b - a [f(x_{0}) + f(x_{n}) + 2f(x_{all else})]

________________________________________________2n

Simpson's Rule: n parabolas (n must be even)

______________2

rough equivalent of b - a [f(x_{0}) + f(x_{n}) + alternate 4s and 2s as coefficients for everything in between]

_________________3n

Calculator: Math 9 f_{n}^{interval} (function, x, 1, 2) Enter

Error Formula:

Trapezoid Rule- E < (b - a)^{3} max |f"|

__________________12n^{2}

_______________________________

[a, b]

Simpson's Rule- E < (b - a)^{5} max |f^{(4)}|

__________________180^{4}

_______________________________

[a, b]

max |f"| = put [a, b] into f" and figure the biggest out; just what it looks like

Resolute Myth
Grey Ultima