Dérivées de fonctions usuelles

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Dérivées de fonctions usuelles
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Leonard Euler
Created by Leonard Euler over 9 years ago
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Resource summary

Question Answer
\[c'\] \[0\]
\[x'\] \[1\]
\[ (x^2)'\] \[2x\]
\[ (x^3)'\] \[3x^2\]
\[ (x^k)'\] \[k\cdot x^{k-1}\]
\[(\frac{1}{x})'\] \[-\dfrac{1}{x^2}\]
\[ (\sqrt{x})'\] \[\dfrac{1}{2\sqrt{x}}\]
\[(\lambda\cdot u)'\] \[\lambda\cdot u'\]
\[(u+v)'\] \[u'+v'\]
\[(u\cdot v)'\] \[u'\cdot v+u\cdot v'\]
\[\Big(\dfrac{1}{u}\Big)'\] \[-\dfrac{u'}{u^2}\]
\[\Big(\dfrac{u}{v}\Big)'\] \[\dfrac{u'\cdot v-u\cdot v'}{v^2}\]
\[(u\circ v)'\] \[(u' \circ v)\cdot v'\]
\[(\sqrt{u})'\] \[\dfrac{u'}{2\sqrt{u}}\]
\[(u^k)'\] \[k\cdot u^{k-1}\cdot u'\]
\[ (\sin{x})'\] \[\cos{x}\]
\[ (\cos{x})'\] \[-\sin{x}\]
\[ (\tan{x})'\] \[1+\tan^2{x}=\dfrac{1}{\cos^2{x}}\]
\[(\sin{u})'\] \[u'\cdot \cos{u}\]
\[(\cos{u})'\] \[-u'\cdot \sin{u}\]
\[(\tan{u})'\] \[u'\cdot(1+\tan^2{u})=\dfrac{u'}{\cos^2{u}}\]
\[ (e^x)'\] \[e^x\]
\[ (ln{x})'\] \[\dfrac{1}{x}\]
\[(e^u)'\] \[u'\cdot e^u\]
\[(\ln{u})'\] \[\dfrac{u'}{u}\]
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