jueves, 17 de junio de 2021

càlcul diferencial y integral

[ES(x)][ f(x)+(-1)·g(x) = d_{x}[ S(x) ] ]

f(x)+(-1)·f(x) = d_{x}[ 1 ]

f(x)+(-1)·g(x) = d_{x}[ P(x) ] <==> g(x)+(-1)·f(x) = d_{x}[ (-1)·P(x) ]

Si f(x)+(-1)·g(x) = d_{x}[ P(x) ] & g(x)+(-1)·h(x) = d_{x}[ Q(x) ] ==>

f(x)+(-1)·h(x) = d_{x}[ P(x)+Q(x) ]


[ES(x)][ f(x)+(-1)·g(x) = int[ S(x) ] d[x] ]

f(x)+(-1)·f(x) = int[ 0^{2} ] d[x]

f(x)+(-1)·g(x) = int[ P(x) ] d[x] <==> g(x)+(-1)·f(x) = int[ (-1)·P(x) ] d[x]

Si f(x)+(-1)·g(x) = int[ P(x) ] d[x] & g(x)+(-1)·h(x) = int[ Q(x) ] d[x] ==>

f(x)+(-1)·h(x) = int[ P(x)+Q(x) ] d[x]


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


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

e^{ln(x)}·d_{x}[ln(x)] = 1

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


f(x) = x^{n}

ln(f(x)) = ln(x^{n}) = n·ln(x)

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


Teoremas:

d_{x}[1] = (1/h)·( (x+h)^{0}+(-1)·x^{0} ) = 0·(h/h) = 0

d_{x}[x] = (1/h)·( (x+h)+(-x) ) = (h/h) = 1

d_{x}[x^{2}] = (1/h)·( x^{2}+2xh+h^{2}+(-1)·x^{2} ) = 2x+h = 2x

d_{x}[x^{n}] = ...

... (1/h)·( x^{n}+nx^{n+(-1)}h+...+h^{n}+(-1)·x^{n} ) = ...

... nx^{n+(-1)}+...+h^{n+(-1)} = nx^{n+(-1)}


d_{x}[sin(x)] = (1/h)·( sin(x+h)+(-1)·sin(x) ) = ...

... (1/h)·( sin(x)cos(h)+cos(x)sin(h)+(-1)·sin(x) ) = cos(x)·( sin(h)/h ) = cos(x)

d_{x}[cos(x)] = (1/h)·( cos(x+h)+(-1)·cos(x) ) = ...

... (1/h)·( cos(x)cos(h)+(-1)·sin(x)sin(h)+(-1)·cos(x) ) = (-1)·sin(x)·( sin(h)/h ) = (-1)·sin(x)


int[ arc-tan(x) ] d[x] = x·arc-tan(x)+(-1)·(1/2)·ln( 1+x^{2} )

int[ arc-cot(x) ] d[x] = x·arc-cot(x)+(1/2)·ln( 1+x^{2} )


int[ arc-sin(x) ] d[x] = x·arc-sin(x)+(-1)·( 1+(-1)·x^{2} )^{(1/2)}

int[ arc-cos(x) ] d[x] = x·arc-cos(x)+( 1+(-1)·x^{2} )^{(1/2)}


int[ x^{n}·arc-sin(x) ] d[x] = ...

... (1/(n+1))·x^{n+1}·arc-sin(x)+...

... ( ( 1+(-1)·x^{2} )^{(1/2)} [o(x)o] ( 1/(n+1)^{2} )·x^{n+1} )

int[ x^{n}·arc-cos(x) ] d[x] = ... 

... (1/(n+1))·x^{n+1}·arc-cos(x)+...

... ( (-1)·( 1+(-1)·x^{2} )^{(1/2)} [o(x)o] ( 1/(n+1)^{2} )·x^{n+1} )

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