8/17/2019 DERIVADAS E INTEGRALES 2016.doc
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Análisis I Tabla-26/C
Instituto de Matemáticas
Tabla de Derivadas e Integrales
Función Derivada Integral
y = c y' = 0 c.xy = c.x y' = c c.x2/2
y = xn y' = n.xn-1 xn+1/n+1
y = x-n y’ = -1/(n.xn-1) x-n+1/-n+1
y = x½ y’ = 1/(2.x½) 2.x3/2/3
y = xa/b y' = a.x(a/b)-1/b x(a/b)+1/[(a/b)+1]
y = 1/x y' = -1/x2 ln x
y = sen x y' = cos x -cos x
y = cos x y' = -sen x sen x
y = tg x y' = 1/cos2x -ln cos x
y = cotg x y' = -1/sen2x ln sen x
y = sec x y' = sen x/cos2x ln (tg ½.x)
y = cosec x y' = -cos x/sen2x ln [cos x/(1 - sen x)]
y = arcsen x y' = 1/(1 - x2)½ x.arcsen x + (1 - x2)½
y = arccos x y' = -1/(1 - x2)½ x.arccos x - (1 - x2)½
y = arctg x y' = 1/(1 + x2) x.arctg x - ½ln (1 + x2)
y = arccotg x y' = -1/(1 + x2) x.arccotg x + ½ln (1 + x2)
y = arcsec x y' = 1/[x.(x2 -1)½] 1
y = arccosec x y' = -1/[x.(x2 – 1)½] 2
y = sen x y' = cos x cos x
y = cos x y' = sen x sen x
y = tg x y' = sec2x ln cos x
y = cotg x y' = -cosec2x ln sen x
y = sec x y' = -sec x.tg x 3
y = cosec x y' = -cosec x.cotg x !
y = ln x y' = 1/x x.(ln x - 1)
y = logax y' = 1/x.ln a x.( logax - 1/ln a)
y = ex y' = ex ex
y = ax y' = ax.ln a ax/ln a
y = xx y' = xx.(ln x + 1) "
y = e# y’ = e#.#’ $
y = #.% y' = #'.% + %'.# ∫ #.&% + ∫ %.
y = #/% y' = (#'.% - %'.#)/%2
y = #% y' = #%.(%'.ln# + %.#'/#)
y = ln#% y’ = (%’.#.ln# - #’.%.ln%)/%.#.ln2#
Instituto de Matemáticas
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