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25 Cards in this Set
- Front
- Back
Rule 1 of Logarithmic Differentiation
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a^b = e^(blna)
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Rule 2 of Logarithmic Differentiation
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d/dx ca^x = ca^x*lna
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L'Hôpital's Rule
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If lim g(x)/h(x) is of an indeterminate form, then lim g(x)/h(x) = lim g'(x)/h'(x)
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CRI (Chain Rule for Integrals)
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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
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catenary
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hanging cable curve -turns out to be a cosh curve!
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algorithm
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a well defined process of multiple steps that produces the answer to some question
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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 |
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∫tanxdx
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-ln|cosx|+C or ln|secx|+C
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∫cscxdx
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-ln|cscx+cotx|+C
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∫secxdx
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ln|secx+tanx|+C
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∫cotxdx
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ln|sinx|+C
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∫lnxdx
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xlnx-x+C
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d/dx (xlnx-x+C)
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lnx
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∫sinhxdx
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coshx+C
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∫coshxdx
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sinhx+C
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∫tanhxdx
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ln|coshx|+C
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∫cothxdx
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ln|sinhx|+C
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Write e^x a complicated way.
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lim as n approaches infnity of (1+x/n)^n
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sinhx (sinch)
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(e^x-e^(-x))/2
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coshx (kahsh)
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(e^x+e^(-x))/2
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tanhx (thhan)
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sinhx/coshx
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cschx (coshek)
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1/sinhx
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sechx (shek)
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1/coshx
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cothx (cothhan)
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1/tanhx
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Simpson's Rule
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1/3(∆x)(Y0+4Y1+2Y2+4Y3+2Y4+...+2Yn-2+4Yn-1+Yn)
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