Lys van afgeleides

Saam met integrasie vorm differensiasie die hoofbewerkings van calculus. 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

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

Kettingreël

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

Afgeleides van eenvoudige funksies

${d \over dx} c = 0$

${d \over dx} x = 1$

${d \over dx} cx = c$

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

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

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

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

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

Afgeleides van eksponensiaalfunksies en logaritmes

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

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

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

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

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

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

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