Deribatu taula

Kalkulu diferentzialean egiten den eragiketarik ohikoenetarikoa da funtzio baten deribatua aurkitzea. Taula honen laguntzaz esker edozein funtzio elementalen deribatua kalkula daiteke. Esan beharra dago, taulako f eta g funtzioak deribagarriak eta c zenbaki errealak direla suposatuko dela.

Funtzio orokorren deribaziorako arauak

Diferentzialaren linealtasuna

$\left({cf}\right)' = cf'$

$\left({f + g}\right)' = f' + g'$

$\left({f - g}\right)' = f' - g'$

Biderkaduraren deribatua

$\left({fg}\right)' = f'g + fg'$

Zatiduraren deribatua

$\left({f \over g}\right)' = {f'g - fg' \over g^2}, \qquad g \ne 0$

Funtzio potentzialaren deribatua

$(f^g)' = \left(e^{g\ln f}\right)' = f^g\left(f'{g \over f} + g'\ln f\right),\qquad f>0$

Funtzio konposatuaren edo katearen erregela

$(f \circ g)' = (f' \circ g)g'$

Logaritmoaren deribatua

$f' = (\ln f)'f, \qquad f>0$

Funtzio sinpleen deribatuak

${d \over dx} c = 0$

${d \over dx} x = 1$

${d \over dx} cx = c$

${d \over dx} |x| = {|x| \over x} = \sgn x,\qquad x \ne 0$

${d \over dx} x^c = cx^{c-1} \qquad \mbox{where both } x^c \mbox{ and } cx^{c-1} \mbox { are defined}$

${d \over dx} \left({1 \over x}\right) = {d \over dx} \left(x^{-1}\right) = -x^{-2} = -{1 \over x^2}$

${d \over dx} \left({1 \over x^c}\right) = {d \over dx} \left(x^{-c}\right) = -{c \over x^{c+1}}$

${d \over dx} \sqrt{x} = {d \over dx} x^{1\over 2} = {1 \over 2} x^{-{1\over 2}} = {1 \over 2 \sqrt{x}}, \qquad x > 0$

Funtzio esponentzial eta logaritmikoen deribatuak

${d \over dx} c^x = {c^x \ln c},\qquad c > 0$

${d \over dx} e^x = e^x$

${d \over dx} \log_c x = {1 \over x \ln c},\qquad c > 0, c \ne 1$

${d \over dx} \ln x = {1 \over x},\qquad x > 0$

${d \over dx} \ln |x| = {1 \over x}$

${d \over dx} x^x = x^x(1+\ln x)$

Funtzio trigonometrikoen deribatuak

${d \over dx} \sin x = \cos x$

${d \over dx} \cos x = -\sin x$

${d \over dx} \tan x = \sec^2 x = { 1 \over \cos^2 x}$

${d \over dx} \sec x = \tan x \sec x$

${d \over dx} \cot x = -\csc^2 x = { -1 \over \sin^2 x}$

${d \over dx} \csc x = -\csc x \cot x$

${d \over dx} \arcsin x = { 1 \over \sqrt{1 - x^2}}$

${d \over dx} \arccos x = {-1 \over \sqrt{1 - x^2}}$

${d \over dx} \arctan x = { 1 \over 1 + x^2}$

${d \over dx} \arcsec x = { 1 \over |x|\sqrt{x^2 - 1}}$

${d \over dx} \arccot x = {-1 \over 1 + x^2}$

${d \over dx} \arccsc x = {-1 \over |x|\sqrt{x^2 - 1}}$

Funtzio hiperbolikoen deribatuak

${d \over dx} \sinh x = \cosh x$

${d \over dx} \cosh x = \sinh x$

${d \over dx} \tanh x = \operatorname{sech}^2\,x$

${d \over dx}\,\operatorname{sech}\,x = - \tanh x\,\operatorname{sech}\,x$

${d \over dx}\,\operatorname{coth}\,x = -\,\operatorname{csch}^2\,x$

${d \over dx}\,\operatorname{csch}\,x = -\,\operatorname{coth}\,x\,\operatorname{csch}\,x$

${d \over dx}\,\operatorname{arcsinh}\,x = { 1 \over \sqrt{x^2 + 1}}$

${d \over dx}\,\operatorname{arccosh}\,x = { 1 \over \sqrt{x^2 - 1}}$

${d \over dx}\,\operatorname{arctanh}\,x = { 1 \over 1 - x^2}$

${d \over dx}\,\operatorname{arcsech}\,x = { -1 \over x\sqrt{1 - x^2}}$

${d \over dx}\,\operatorname{arccoth}\,x = { 1 \over 1 - x^2}$

${d \over dx}\,\operatorname{arccsch}\,x = {-1 \over |x|\sqrt{1 + x^2}}$

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