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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)