CALCULUS I TOOLBOX

1) INEQUALITIES:

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.

2) LIMITS:

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:

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3) DERIVATIVES:

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'.

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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:

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c) M2I2ACIDS

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)).

8) ANTIDIFFERENTIATION:

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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