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Función Derivada Integralf ( x )=k f ´ ( x )=0conk=cte ∫ k dx=kxf ( x )=x f ´ ( x )=1 ∫ x dx= x2

2f ( x )=1 f ´ ( x )=0 ∫ dx=xf ( x )=xn f ´ ( x )=nxn−1

∫ xn dx= xn+1

n+1conn≠1

f ( x )= n√x f ´ ( x )= 1n n√ xn−1 ∫ n√x dx= x

1n+1

1n+1

f ( x )=k .g( x) f ´ ( x )=k .g ´ ( x )conk=cte ∫ k g (x) = k∫ g ( x ) dx

f ( x )=[u±v ] f ´ ( x )=u´ ±v ´ ∫u± v dx = ∫udx ±∫ v dxf ( x )=[u . v ] f ´ ( x )=u ´ . v+u . v ´ --------------------

f ( x )=[u . v .w ] f ´ ( x )=u´ . v .w+u . v ´ .w+u . v .w ´ --------------------

f ( x )=[uv ] f ´ ( x )=u ´ .v−u . v ´v2

--------------------

f ( x )= 1x

f ´ ( x )=−1x2

∫ 1x dx=ln|x|

f ( x )=√x f ´ ( x )= 12√x

f ( x )=ln (x ) f ´ ( x )=1x

∫ ln (x )dx=x . ln ( x )−x

f ( x )=ex f ´ ( x )=ex ∫ ex dx=ex

f ( x )=e−x f ´ ( x )=−e−x ∫ e−x dx=−e− x

2

f ( x )=ax f ´ ( x )=ax . ln (a) ∫ ax dx= ax

ln (a)f ( x )=x x f ´ ( x )=x x . ¿

ln (x)¿No tiene primitiva

f ( x )=sen (x) f ´ ( x )=cos (x ) ∫ sen(x )dx=−cos (x)f ( x )=cos (x ) f ´ ( x )=−sen (x) ∫cos (x )dx=sen (x)f ( x )=tg(x ) f ´ ( x )=sec2 (x ) ∫ tg (x )dx=−ln (cos (x))

f ( x )=cosec (x) f ´ ( x )=−cosec ( x ) . cotg (x) ∫ cosec(x)dx=− ln (cosec ( x )+cotg(x ))f ( x )=sec (x ) f ´ ( x )=sec ( x ) . tg(x) ∫ sec(x)dx=ln (sec (x )+tg (x ))f ( x )=cotg (x) f ´ ( x )=cosec2 (x) ∫ cotg( x)dx=ln|sen(x )|

f ( x )=arc sen (x) f ´ ( x )= 1√1−x2

∫ arc sen (x)dx=x .arc sen ( x )+√1−x2

f ( x )=arc cos( x) f ´ ( x )= −1√1−x2

∫ arc cos(x )dx=x .arc cos (x )−√1−x2

f ( x )=arc tg (x) f ´ ( x )= 11+x2

∫ arc tg( x)dx=x .arc tg ( x )−ln (√1+x2)

f ( x )=arc csc(x ) f ´ ( x )= −1|x|√1−x2

∫ arc csc(x)dx=x .arc csc ( x )+ ln|x+√x2−1|

f ( x )=arc sec(x ) f ´ ( x )= 1|x|√1−x2

∫ arc sec(x )dx=x .arc sec ( x )− ln|x+√x2−1|

f ( x )=arc cotg(x) f ´ ( x )= −11+x2

∫ arc cotg(x)dx=x .arc cotg ( x )+ln (√1−x2)

f ( x )=senh (x) f ´ ( x )=cosh (x) ∫ senh(x )dx=cosh (x )f ( x )=cosh (x) f ´ ( x )=senh( x) ∫cosh (x )dx=senh (x )f ( x )=tgh(x ) f ´ ( x )=sech2 (x) ∫ tgh (x)dx=ln (cosh (x))

f ( x )=cosech(x ) f ´ ( x )=−cosec h (x ) . cotgh (x) ∫ cosech(x )dx=−ln (cosech ( x )+cotgh (x))

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f ( x )=sech (x) f ´ ( x )=−sec h ( x ) . tgh( x) ∫ sech( x)dx=2arc tg (ex ) ó −2arc tg(e− x)f ( x )=cotgh(x ) f ´ ( x )=−cosech2 ( x) ∫ cotgh(x )dx=ln (senh ( x ))

f ( x )=arg senh(x )f ´ ( x )= 1

√ x2+1

Hiperbólicas Inversas

∫ 1√a2−x2

dx=arco sen ( xa )

f ( x )=argcosh (x ) f ´ ( x )= 1√ x2−1

∫ 1a2+x2

dx=1a

arco sen ( xa )

f ( x )=argtgh(x) f ´ ( x )= 11−x2 ∫ 1

x .√ x2−a2dx=1

aarco sec( x

a )f ( x )=arg csch(x ) f ´ ( x )= −1

|x|√x2+1∫ 1

√ x2± a2dx=ln (x+√ x2±a2¿)¿

f ( x )=arg sech( x) f ´ ( x )= −1x √1−x2

∫ 1a2−x2

dx= 12aln|a+x

a−x|f ( x )=arg cotgh(x ) f ´ ( x )= 1

1−x2 ∫ 1x .√a2±x2

dx=−1a ln( a+√a2± x2

|x| )

Integración de funciones pares e impares Integración por partes

Si f(x) es par ∫−a

a

f (x ) dx=2∫0

a

f (x)dx ∫u . v dx=u . v−∫ v du

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Si f(x) es impar ∫−a

a

f (x ) dx=0

Área entre 2 curvas

A=∫a

b

[ f ( x )−g(x )]

∫0

a

f ( x )dx=¿−∫0

−a

f (−x ) dx¿

Relaciones Pitagóricas

sen2 (x )+¿ cos2 ( x )=1 // cos2 ( x )=1−sen2 ( x ) // sen2 (x )=1−cos2 ( x )

tg2 ( x )+1=sec2 (x )

cotg2 ( x )+1=cosec2 (x )

c2=a2+b2

Función de la suma o la diferencia

cos ( a+b )=cos (a ) .cos (b )−sen (a ) . sen (b) sen (a )+sen (b )=2. sen( a+b2 ) .cos ( a−b

2 )

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cos ( a−b )=cos (a ) .cos (b )+sen (a ) . sen (b) sen (a )−sen (b )=2. cos( a−b2 ) . sen ( a−b

2 )sen (a+b )=sen (a ) .cos (b )+cos (a ) . sen (b) cos(a )+cos (b )=2. cos(a+b

2 ) .cos ( a−b2 )

sen (a−b )=sen (a ) .cos (b )−cos (a ) . sen(b) cos(a )−cos (b )=2. sen ( a+b2 ) . sen ( a−b

2 )tg (a+b )=

tg (a )+tg (b)1−tg ( a ) .tg(b)

sen (a+b )+sen (a−b )=2. sen (a ) .cos (b)

tg (a−b )=tg (a )−tg(b)1+tg ( a ) .tg(b)

sen (a−b )−sen (a−b )=2. cos ( a ) . sen (b)

Función de ángulo duplo y medio

Función de ángulo duplo

sen (2.a )=2. sen (a ) .cos (a )

cos (2.a )=cos2 (a )−sen2(a)

tg (2.a )= 2. tg(a)1−tg2(a)

=sen (a)

(1+cos ( a ))

Función de ángulo medio

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sen(a2 )=±√ 1−cos (a)

2

cos (a2 )=±√ 1+cos (a)

2

tg( a2 )=±√ 1−cos (a)

1+cos (a)

Función del producto

cos2 (a )=1+cos (2.a)2

sen2 (a )=1−cos (2.a)2

tg2 ( a )=1−cos (2.a)1+cos (2.a)

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sen (a ) .cos (b)=12 [sen (a+b )+sen(b−a)]

cos ( a ) . sen (b )=12 [sen (a+b )−sen (b−a ) ]

cos ( a ) .cos (b)=12 [cos (a+b )+cos(b−a)]

sen (a ) . sen (b)=12 [cos (a−b )−cos (b+a)]

Funciones Hiperbólicas

senh ( x )= ex−e− x

2

cosh ( x )= ex+e−x

2

tgh ( x )= ex−e−x

e x+e−x

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cosech (x )= 1senh(x )

= 2ex−e−x

sech ( x )= 1cosh (x)

= 2ex+e−x

cotgh (x )= 1tgh (x)

= e x+e− x

ex−e−x

arg senh (x )=ln (x+√1+x2 )

arg co sh ( x )=ln (x+√ x2−1)con x≥1

arg tgh ( x )=12ln|1+x1−x|con|x|<1

cosh2 (x )−senh2 ( x )=1 //cosh2 (x )=1+senh2 ( x ) // senh2 (x )=−1+cosh2(x)

cosh ( x )+senh ( x )=e x // senh ( x )=ex−cosh ( x ) // cosh ( x )=ex−senh(x )

cosh ( x )−senh (x )=e−x // senh ( x )=e−x−cosh ( x ) // cosh ( x )=e−x−senh (x)

senh ( x+ y )=senh (x ) . senh ( y )+cosh ( x ) . senh ( y )

cosh ( x+ y )=cosh (x ) . cosh ( y )+senh ( x ) . senh( y )

senh (2x )=2 senh ( x ) .cosh (x )

cosh (2x )=cosh2 ( x )+senh2(x)

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senh2 (x )=cosh (2x )−12

cosh2 (x )=1+cosh (2x )2

Ecuación de la recta tangente

y=f ´ (a ) . ( x−a )+ f (a ) En x=a

Ecuación de la recta secante

y=y2− y1x2−x1

( x−x1 )+ y1 Dadados puntos

y=m ( x−x1 )+ y1Dada la pendiente(m) y un punto

Función cuadrática

y=ax2+bx+c Forma Explícita .

y=a(x−xv )2+ yv FormaCanónica (necesito conocer el vertice ) xv=

−b2a

; despuesreemplazo en la funcion y obtengo yv .

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y=a (x−x1) ( x− x2 ) Forma Factorizada (neceesitoconocer las raices ) x1 ;2=−b±√b2−4 ac

2a