by Eddie Pitch August 26, 2010
A term of endearment used between friends that implies the person has been involved in something mischievous.
by C. Aleccia February 26, 2008
by KROQKEN November 18, 2011
People who refuse to believe in this age of cellphones that there are people who cannot hear their phone rings or is constantly out of reception for more than 5 seconds, and will call nonstop until the other party picks up the phone.
by Jason Wang March 26, 2004
by Countyeasy June 05, 2017
In calculus, the chain rule is used to differentiate compositions of functions. It states that for any function f which is dependent on a variable u, and u is a function of a second variable x, then f is a function of x.
In Set Theory:
f(u)=u V u(x)=x V Vf(x)
In Calculus:
df/dx = du/dx(dx/du)
In Set Theory:
f(u)=u V u(x)=x V Vf(x)
In Calculus:
df/dx = du/dx(dx/du)
to differentiate (x+1)^2, one could multiply this out and apply the sum rule (derivative of a sum is the sum of the derivatives); but what if it were (x+1)^55? Suddenly the multipling out doesn't look so nice. To apply the chain rule, we must first define the functions. Let f(x)=(x+1)^55 and u=(x+1). Therefore f(x)=u^55. By the chain rule:
df/dx = df/du(du/dx)
df/dx = d/du(u^55)(du/dx)
df/dx = 55u(du/dx)
We earlier defined the variable "u" as (x+1). Now we substitute this in.
df/dx = 55(x+1)d/dx(x+1)
df/dx = 55x+55(1)
df/dx = 55x+55
df/dx = df/du(du/dx)
df/dx = d/du(u^55)(du/dx)
df/dx = 55u(du/dx)
We earlier defined the variable "u" as (x+1). Now we substitute this in.
df/dx = 55(x+1)d/dx(x+1)
df/dx = 55x+55(1)
df/dx = 55x+55
by Gladwarez July 11, 2006
Wild; out of control
etym.: possibly derived from Fr. 'se déchaîner,' to burst out; to go wild; to arouse
etym.: possibly derived from Fr. 'se déchaîner,' to burst out; to go wild; to arouse
by lairdweller October 18, 2009