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operador médico

operador médico constructor asterisco


{1,i,(-1),(-i)}


{k,(-j),(-k),j}


(-j)   i    k
(-1)  0   1
(-k) (-i)  j

operaciones médicas

operador star-trek-Pável-Checkov


operador constructor
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operador destructor
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operador constructor
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operador destructor
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geofísica electro-exponencial

E(x,y,z) = qk·< ( (e^{ln(a)·x}+(-1))/x ) , ( (e^{ln(a)·y}+(-1))/y ) , ( (e^{ln(a)·z}+(-1))/z ) >


E(0,0,0) = qk·< ln(a) , ln(a) , ln(a) >


div[ E(x,y,z) ] =  qk·( ...
... ( (e^{ln(a)·x}+(-1))/x )·ln(a)+( ln(a)/x )+(-1)·( (e^{ln(a)·x}+(-1))/x^{2} ) +..
... ( (e^{ln(a)·y}+(-1))/y )·ln(a)+( ln(a)/y )+(-1)·( (e^{ln(a)·y}+(-1))/y^{2} ) +...
... ( (e^{ln(a)·z}+(-1))/z )·ln(a)+( ln(a)/z )+(-1)·( (e^{ln(a)·z}+(-1))/z^{2} ) ...
... )


div[ E(0,0,0) ] = 3qk·( ln(a) )^{2}

criteri de derivades

lim[h-->0][ (h/h)·( f(x+h)+(-1)f(x) )/( g(x+h)+(-1)g(x) ) ] = lim[h-->0][ ( f(x+h) )/( g(x+h) ) ] = ...
... ( f(x)/g(x) ) <==> ( f(x) = 0  & g(x) = 0 )

derivada de la exponencial

d_{x}[e^{x}] = lim[h-->0][ (1/h)( e^{x+h}+(-1)·e^{x} ) ]
d_{x}[e^{x}] = e^{x}·lim[h-->0][ (1/h)( e^{h}+(-1) ) ]
d_{x}[e^{x}] = e^{x}·lim[h-->0][ (1/h)( (1+h+(1/2)·h^{2}+...(1/n!)·h^{n}+...)+(-1) ) ]
d_{x}[e^{x}] = e^{x}


d_{x}[a^{x}] = lim[h-->0][ (1/h)( a^{x+h}+(-1)·a^{x} ) ]
d_{x}[a^{x}] = a^{x}·lim[h-->0][ (1/h)( a^{h}+(-1) ) ]
d_{x}[a^{x}] = a^{x}·lim[h-->0][ (1/h)( e^{ln(a)·h}+(-1) ) ]
d_{x}[a^{x}] = a^{x}·lim[h-->0][ (1/h)( (1+(ln(a)·h)+(1/2)·(ln(a)·h)^{2}+...(1/n!)·(ln(a)·h)^{n}+...)+(-1) ) ]
d_{x}[a^{x}] = a^{x}·ln(a)

derivada del logaritme

d_{x}[x] = 1
d_{x}[e^{ln(x)}] = 1
d_{ln(x)}[e^{ln(x)}]·d_{x}[ln(x)] = 1
e^{ln(x)}·d_{x}[ln(x)] = 1
x·d_{x}[ln(x)] = 1
d_{x}[ln(x)] = (1/x)


d_{x}[x+a] = 1
d_{x}[e^{ln(x+a)}] = 1
d_{ln(x+a)}[e^{ln(x+a)}]·d_{x}[ln(x+a)] = 1
e^{ln(x+a)}·d_{x}[ln(x+a)] = 1
(x+a)·d_{x}[ln(x+a)] = 1
d_{x}[ln(x+a)] = ( 1/(x+a) )


d_{x}[x^{n}] = nx^{(n+(-1))}
d_{x}[e^{ln(x^{n})}] = nx^{(n+(-1))}
d_{ln(x^{n})}[e^{ln(x^{n})}]·d_{x}[ln(x^{n})] = nx^{(n+(-1))}
e^{ln(x^{n})}·d_{x}[ln(x^{n})] = nx^{(n+(-1))}
x^{n}·d_{x}[ln(x^{n})] = nx^{(n+(-1))}
d_{x}[ln(x^{n})] = (1/x^{n})·nx^{(n+(-1))}
d_{x}[ln(x^{n})] = (n/x)