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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 x-values
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
-ln|cosx|+C or ln|secx|+C
∫cscxdx
-ln|cscx+cotx|+C
∫secxdx
ln|secx+tanx|+C
∫cotxdx
ln|sinx|+C
∫lnxdx
xlnx-x+C
d/dx (xlnx-x+C)
lnx
∫sinhxdx
coshx+C
∫coshxdx
sinhx+C
∫tanhxdx
ln|coshx|+C
∫cothxdx
ln|sinhx|+C
Write e^x a complicated way.
lim as n approaches infnity of (1+x/n)^n
sinhx (sinch)
(e^x-e^(-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+...+2Yn-2+4Yn-1+Yn)