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:<math> \left( x^x \right)' = x^x(1+\ln x)</math>
JANE BAZE PER TE MESUAR ANALIZEN MATEMATIKE
== Derivatet e funksioneve trigonometrike ==
{| style="width:100%; background:transparent; margin-left:2em;"
|width=50%|<math> (\sin x)' = \cos x \,</math>
|width=50%|<math> (\arcsin x)' = { 1 \over \sqrt{1 - x^2}} \,</math>
|-
|<math> (\cos x)' = -\sin x \,</math>
|<math> (\arccos x)' = {-1 \over \sqrt{1 - x^2}} \,</math>
|-
|<math> (\tan x)' = \sec^2 x = { 1 \over \cos^2 x} \,</math>
|<math> (\arctan x)' = { 1 \over 1 + x^2} \,</math>
|-
|<math> (\sec x)' = \sec x \tan x \,</math>
|<math> (\arcsec x)' = { 1 \over |x|\sqrt{x^2 - 1}} \,</math>
|-
|<math> (\csc x)' = -\csc x \cot x \,</math>
|<math> (\arccsc x)' = {-1 \over |x|\sqrt{x^2 - 1}} \,</math>
|-
|<math> (\cot x)' = -\csc^2 x = { -1 \over \sin^2 x} \,</math>
|<math> (\arccot x)' = {-1 \over 1 + x^2} \,</math>
|}
== Derivatet e funksioneve hiperbolike ==
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