Leerstof onder andere voor:[br][math]\oplus[/math]ingangsexamen geneeskunde

[math][br]\begin{tabular}{lll}[br](af+bg)'(x)=af'(x)+bg'(x)\;\; (a,b\in\mathbb{R})&&(\square+\triangle)'=\square'+\triangle' \\[br](f\cdot g)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x) &&(\square\cdot\triangle)'=\square'\cdot\triangle+\square\cdot\triangle' \\[br]\left(f^r\right)'(x)=rf^{r-1}(x)\cdot f'(x) \;\;(r\in \mathbb{R})&&\left(\square^r\right)'=r\square^{r-1}\cdot \square' \\[br]\left(\frac{1}{f}\right)'=\frac{-f'(x)}{f(x)^2}&&\left(\frac{1}{\square}\right)'=\frac{-\square'}{\square^2}\\[br]\left(\frac{f}{g}\right)'(x)=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}&&\left(\frac{\square}{\triangle}\right)'=\frac{\square'\cdot \triangle-\square \cdot \triangle'}{\triangle^2}\\[br]\left(f(g(x)\right)'=f'(g(x))\cdot g'(x)&&\\[br]\left(f^{-1}(x)\right)'=\frac{1}{f'(f^{-1}(x))}&&[br]\end{tabular}[br][/math]

[math][br]\begin{tabular}{lll}[br]f(x)=c\; \left(c\in\mathbb{R}\right)&f'(x)=0&\\[br]f(x)=x&f'(x)=1&\\[br]f(x)=x^r\;(r\in\mathbb{R})&f'(x)=rx^{r-1}&\left(\square^r\right)'=r\cdot\square^{r-1}\cdot\square'\\[br]f(x)=\sqrt{x}&f'(x)=\frac{1}{2\sqrt{x}}&\left(\sqrt{\square}\right)'=\frac{1}{2\sqrt{\square}}\cdot\square'\\[br]f(x)=e^x&f'(x)=e^x&\left(e^{\square}\right)'=e^{\square}\cdot\square'\\[br]f(x)=a^x\;\left(a\in\mathbb{R}_{0}^{+}\backslash\{1\}\right)&f'(x)=a^x lna&\left(a^{\square}\right)'=a^{\square}lna\cdot\square'\\[br]f(x)=lnx&f'(x)=\frac{1}{x}&\left(ln\square\right)'=\frac{1}{\square}\cdot\square'\\[br]f(x)=^{a}\log x&f'(x)=\frac{1}{x\ln a}&\left(^{a}\log\square\right)'=\frac{1}{\square\ln a}\cdot\square'\\[br]f(x)=\sin x&f'(x)=\cos x& \left(sin\square\right)'=\cos\square\cdot\square'\\[br]f(x)=\cos x & f'(x)=-\sin x&\left(\cos\square\right)'=-\sin\square\cdot\square'\\[br]f(x)=\tan x & f'(x)=\frac{1}{\cos^2 x}&\left(\tan \square\right)'=\frac{1}{cos^2\square}\cdot\square'\\[br]f(x)=\cot x & f'(x)=\frac{-1}{\sin^2 x}& \left(\cot\square\right)'=\frac{-1}{sin^2\square}\cdot\square'\\[br]f(x)=\arcsin x & f'(x)=\frac{1}{\sqrt{1-x^2}}&\left(\arcsin \square\right)'=\frac{1}{\sqrt{1-\square^2}}\cdot\square'\\[br]f(x)=\arccos x & f'(x)=\frac{-1}{\sqrt{1-x^2}}&\left(\arccos \square\right)'=\frac{-1}{\sqrt{1-\square^2}}\cdot\square'\\[br]f(x)=\mbox{arc}\tan x&f'(x)=\frac{1}{1+x^2}&\left(\mbox{arc}\tan\square\right)'=\frac{1}{1+\square^2}\cdot\square'\\[br]f(x)=\mbox{arc}\cot x&f'(x)=\frac{-1}{1+x^2}&\left(\mbox{arc}\cot\square\right)'=\frac{-1}{1+\square^2}\cdot\square'\\[br]\end{tabular}[br][/math]

Op de volgende sites kan je online oefenen.[br][url]http://www.bluffton.edu/homepages/facstaff/nesterd/java/derivs.html[/url][br][url]http://www.tkiryl.com/Tests/[/url]

Op deze site kan je jouw antwoord controleren:[br][url]https://www.derivative-calculator.net/[/url]