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25 Cards in this Set
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Rule 1 of Logarithmic Differentiation

a^b = e^(blna)


Rule 2 of Logarithmic Differentiation

d/dx ca^x = ca^x*lna


L'Hôpital's Rule

If lim g(x)/h(x) is of an indeterminate form, then lim g(x)/h(x) = lim g'(x)/h'(x)


CRI (Chain Rule for Integrals)

d/dx ∫ from c to v(x) of f(t)dt = f(v(x))*v'(x) provided f is continuous on interval [c,v(x)] and v differentiable on at least the domain of xvalues


catenary

hanging cable curve turns out to be a cosh curve!


algorithm

a well defined process of multiple steps that produces the answer to some question


What is a solution to a diffeq?
(General vs. Particular) How do we prove it works? 
An equation specifying a relation (or more preferably a function) that satisfies 1. The differential equation 2. The initial conditions, if any.
a GENERAL equation satisfies the first bullet a PARTICULAR equations satisfies both To prove it works, 1. Take the derivative 2. Plug in to check initial conditions 

∫tanxdx

lncosx+C or lnsecx+C


∫cscxdx

lncscx+cotx+C


∫secxdx

lnsecx+tanx+C


∫cotxdx

lnsinx+C


∫lnxdx

xlnxx+C


d/dx (xlnxx+C)

lnx


∫sinhxdx

coshx+C


∫coshxdx

sinhx+C


∫tanhxdx

lncoshx+C


∫cothxdx

lnsinhx+C


Write e^x a complicated way.

lim as n approaches infnity of (1+x/n)^n


sinhx (sinch)

(e^xe^(x))/2


coshx (kahsh)

(e^x+e^(x))/2


tanhx (thhan)

sinhx/coshx


cschx (coshek)

1/sinhx


sechx (shek)

1/coshx


cothx (cothhan)

1/tanhx


Simpson's Rule

1/3(∆x)(Y0+4Y1+2Y2+4Y3+2Y4+...+2Yn2+4Yn1+Yn)
