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

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

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

${d \over dx}c=0$

${d \over dx}x=1$

${d \over dx}cx=c$

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

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

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

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

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

Funtzio esponentzial eta logaritmikoen deribatuak

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

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

${d \over dx}\log _{c}x={1 \over x\ln c},\qquad c>0,c\neq 1$

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

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

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

Funtzio trigonometrikoen deribatuak

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

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

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

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

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

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

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

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

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

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

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

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

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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