|CALCULUS I TOOLBOX|
basic rule: reverse the direction of the inequality when multiplying or dividing by a negative.
a) absolute value: | f(x) | < b becomes -b < f(x) < b (sandwich)
b) quadratics: relate to zero, locate zeroes, then solve with a number line.
c) fractions: denominator a constant, multiply through by the lowest common multiple of the denominators and solve as a linear inequality.
denominator some f(x) , combine the fractions and relate to zero. Then solve as a quadratic inequality.
a) general limits: lim xd a f(x): factor, cancel and substitute x = a.
|b) trig limits: limhd0 = 1||limhd0 0|
c) limit theorem: lim xd a
d) limits involving infinity:
a) power rule: if f(x) = axn then f '(x) = naxn-1
b) product rule: (uv)' = u'v + v'u, where u and v are fuctions of x.
c) quotient rule: (u/v)' = (u'v - v'u)/ v2
d) chain rule: (f[g(x)])' = f ' [g(x)]g'(x)
e) implicit: assume that y = f(x) exists and use the chain rule on that function to find y'.
f) trig functions:
|(sin u)' = du cos u||(cos u)' = -du sin u|
|(tan u)' = du sec2 u||(cot u)' = -du csc2 u|
|(sec u)' = du sec u tan u||(csc u)' = -du csc u cot u|
Note1: If the function's name includes the "co" syllable (ex: cotan), the derivative is negative!
Note2: Notice the symmetry of the pairs.
For instance, (tan u)' = du sec2 u, so (cot u)' = -du csc2 u
It's the same for the other 2 pairs. Just add the minus sign and the "co" syllable.
g) inverse trig functions:
|(arcsin u)' =||(arccos u)' =|
|(arctan u)' =||(arcsec u)' =|
h) exponential and log functions:
|(eu)' = du eu||(au)' = du au ln a|
|(ln u)' =||(logau)' =|
i) increments and differentials:
y = f(x) u dy = f '(x) dx u y2 = f(x) + f '(x) dx
4) GRAPHING TECHNIQUES:
a) first derivative test:
b) second derivative test:
max, min, intercepts, inflection points, asymptotes, concave up, concave down, increasing, decreasing, and symmetry.
* see hand-out titled "NOTES ON GRAPHS".
5) RELATED RATES PROBLEMS:
6) MAX-MIN PROBLEMS:
7) MEAN VALUE THEOREM:
If f(x) is continuous on [a,b] and differentiable on ]a,b[,
then there exists a "c" on ]a,b[ such that f'(c) =
which means that the slope of the tangent at (c, f(c)) is equal to
the slope of the chord joining (a, f(a)) to (b, f(b)).
a) power rule: if f '(x) = axn then f(x) = + C
b) differential equations: Antidifferentiate until you reach f(x), plugging in boundary values.
f ' ' '(x) d f '' (x) d f '(x) d f(x)
Use boundary values to establish the values of C at each level of antidifferentiation.
Use different letters to indicate the constant in each level of derivative.
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