Lys van afgeleides

Saam met integrasie vorm differensiasie die hoofbewerkings van kalkulus. In die onderstaande lys is f en g differensieerbare funksies van die reële getal s. c is ook 'n reële getal.

Hierdie lys van afgeleides is voldoende om enige elementêre funksie te differensieer.

Algemene reëls by die afleiding van funksies

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

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

Produkreël

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

Kwosiëntreël

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

Kettingreël

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

Afgeleides van eenvoudige funksies

${d \over dx}c=0$

${d \over dx}x=1$

${d \over dx}cx=c$

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

${d \over dx}x^{c}=cx^{c-1}\qquad {\mbox{met beide }}x^{c}{\mbox{ en }}cx^{c-1}{\mbox{ gedefinieer}}$

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

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

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

Afgeleides van eksponensiële funksie en logaritmes

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

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

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

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

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

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

Afgeleides van trigonometriese funksies

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

${d \over dx}{\mbox{bgsin}}x={1 \over {\sqrt {1-x^{2}}}}$ ${d \over dx}{\mbox{bgcos}}x={-1 \over {\sqrt {1-x^{2}}}}$ ${d \over dx}{\mbox{bgtan}}x={1 \over 1+x^{2}}$ ${d \over dx}{\mbox{bgsec}}x={1 \over |x|{\sqrt {x^{2}-1}}}$ ${d \over dx}{\mbox{bgcot}}x={-1 \over 1+x^{2}}$ ${d \over dx}{\mbox{bgcsc}}x={-1 \over |x|{\sqrt {x^{2}-1}}}$

Afgeleides van hiperboliese funksies

${d \over dx}\sinh x=\cosh x={\frac {e^{x}+e^{-x}}{2}}$ ${d \over dx}\cosh x=\sinh x={\frac {e^{x}-e^{-x}}{2}}$ ${d \over dx}\tanh x=\operatorname {sech} ^{2}\,x$ ${d \over dx}\,\operatorname {sech} \,x=-\tanh x\,\operatorname {sech} \,x$ ${d \over dx}\,\operatorname {coth} \,x=-\,\operatorname {csch} ^{2}\,x$ ${d \over dx}\,\operatorname {csch} \,x=-\,\operatorname {coth} \,x\,\operatorname {csch} \,x$

${d \over dx}\,{\mbox{sinh}}^{-1}\,x={1 \over {\sqrt {x^{2}+1}}}$ ${d \over dx}\,{\mbox{cosh}}^{-1}\,x={1 \over {\sqrt {x^{2}-1}}}$ ${d \over dx}\,{\mbox{tanh}}^{-1}\,x={1 \over 1-x^{2}}$ ${d \over dx}\,{\mbox{sech}}^{-1}\,x={-1 \over x{\sqrt {1-x^{2}}}}$ ${d \over dx}\,{\mbox{coth}}^{-1}\,x={1 \over 1-x^{2}}$ ${d \over dx}\,{\mbox{csch}}^{-1}\,x={-1 \over |x|{\sqrt {1+x^{2}}}}$

Afgeleides van inverse funksies

${d \over dx}(f^{-1}(x))={\frac {1}{f'(f^{-1}(x))}}$

Verwysings

1. Stewart, J. (2003). Single Variable Calculus. (5th ed.). Belmont, USA: Thomson Learning.
2. Groenewald, G.J., Hitge, M. (2005). Analise II Studiegids vir WISK121A. Potchefstroom: Noordwes-Universiteit.
3. Jordan, D.W., Smith, P. (2002). Mathematical techniques: An introduction for the engineering, physical and mathematical sciences. USA: Oxford University Press.

Aantekeninge

• Sien ook Lys van integrale

• Hierdie artikel is 'n vertaling van die Engelse Wikipedia artikel Table of derivatives