Derivadas

Cálculo Diferencial

Regla general de derivación:

dy f ( x + ∆x ) − f ( x ) = lim dx ∆x →0 ∆x
FÓRMULAS FUNDAMENTALESPARA LA DERIVACIÓN DE ALGUNAS FUNCIONES:

d (c ) =0 ( c = constante) dx d (x ) =1 dx d (u ± v ± ... ± w) = du ± dv ± ... ± dwdx dx dx dx d (cu ) du =c dx dx d (uv ) dv du =u +v dx dx dx
d xn = nx n −1 dx d un du = nu n −1 dx dx d  u  1  du dv  −u   = 2 v dx  v  v  dx dx 

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d u du e = eu dx dx d v du dv + u v ln u u = vu v −1 dx dx dx dy dy dv(Regla de la cadena) = ⋅ dx dv dx d (sen u ) = cos u du dx dx d (cos u ) = − sen u du dx dx d (tg u ) = sec2 u du dx dx

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d (ctg u ) = − csc2 u du dx dx

d (sec u ) = sec u ⋅ tg u du dx dx d (csc u ) = − csc u ⋅ ctg u du dx dx d du 1 sen−1 u = 2 dx 1 − u dx

d  u  1 du  = dx  c  c dx
d dx

( u ) = 2 1u du dx

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d dx

m u d (uvz ) = uv dz+ uz dv + vz du dx dx dx dx d (ln u ) = 1 du dx u dx

( u )=
m

1
m m −1

du dx

d (log u ) = log e du dx u dx d udu a = a u ln a dx dx

du d 1 cos −1 u = − 2 dx 1 − u dx d 1 du tg −1 u = dx 1 + u 2 dx d 1 du ctg −1 u = − dx 1 + u 2 dx d 1du sec −1 u = dx u u 2 − 1 dx

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du d 1 csc −1 u = − 2 dx u u − 1 dx

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