Analyse/Differentiatie

Afgeleide van een Constante

Voor iedere constante $c$ geldt dat de afgeleide $c'=0$:

$f(x)=c\quad geeft\quad f'(x)=0\!$

Bewijs

${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {c-c}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {0}{\Delta x}}\\&=0&\Box \end{aligned}}$

Afgeleide van een Machtsfunctie

Voor iedere functie $g(x)=ax^{n}(n\in \mathbb {R} )$ geldt:

$g'(x)=nax^{n-1}\!$

Bewijs

Voor gehele, positieve $n$:

Voor dit bewijs gebruiken we het Binomium van Newton:

${\begin{aligned}g'(x)&=\lim _{\Delta x\to 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {a(x+\Delta x)^{n}-ax^{n}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {a{\Big (}\sum _{k=0}^{n}{\binom {n}{k}}x^{k}(\Delta x)^{n-k}{\Big )}-ax^{n}}{\Delta x}}\\&=\lim _{\Delta x\to 0}a{\frac {\sum _{k=0}^{n}{\binom {n}{k}}x^{k}(\Delta x)^{n-k}-x^{n}}{\Delta x}}\\&=\lim _{\Delta x\to 0}a{\frac {\sum _{k=0}^{n-1}{\binom {n}{k}}x^{k}(\Delta x)^{n-k}}{\Delta x}}\\&=\lim _{\Delta x\to 0}a{\Big (}\sum _{k=0}^{n-1}{\binom {n}{k}}x^{k}(\Delta x)^{n-k-1}{\Big )}\\&=a{\binom {n}{n-1}}x^{n-1}\\&=anx^{n-1}&\Box \end{aligned}}$

Voor elke $n\in \mathbb {R}$:

Nog niet geplaatst.

Afgeleide van een Exponentiële Functie

$f(x)=a^{x}\quad geeft\quad f'(x)=a^{x}\cdot \ln(a)$

Vaak wordt de afgeleide van $e^{x}$ (ook: $\exp(x)$) apart vermeld:

$g(x)=e^{x}\quad geeft\quad g'(x)=e^{x}\cdot \ln(e)=e^{x}$

Bewijs

${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {a^{x+\Delta x}-a^{x}}{\Delta x}}\\&=\lim _{\Delta x\to 0}a^{x}{\frac {a^{\Delta x}-1}{\Delta x}}\\&=a^{x}\cdot \lim _{\Delta x\to 0}{\frac {a^{\Delta x}-1}{\Delta x}}\\&=a^{x}\cdot \ln(a)&\Box \end{aligned}}$

Somregel

$h(x)=f(x)+g(x)\quad geeft\quad h'(x)=f'(x)+g'(x)$

Bewijs

Als $h(x)=f(x)+g(x)$, dan geldt:

${\begin{aligned}h'(x)&=\lim _{\Delta x\to 0}{\frac {h(x+\Delta x)-h(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\Big (}f(x+\Delta x)+g(x+\Delta x){\Big )}-{\Big (}f(x)+g(x){\Big )}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)+g(x+\Delta x)-f(x)-g(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)+g(x+\Delta x)-g(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}\left({\frac {f(x+\Delta x)-f(x)}{\Delta x}}+{\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right)\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}+\lim _{\Delta x\to 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}\\&=f'(x)+g'(x)&\Box \end{aligned}}$

Productregel

$h(x)=f(x)\cdot g(x)\quad geeft\quad h'(x)=f'(x)g(x)+f(x)g'(x)$

Bewijs

Als $h(x)=f(x)\cdot g(x)$, dan geldt:

${\begin{aligned}h'(x)&=\lim _{\Delta x\to 0}{\frac {h(x+\Delta x)-h(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}\left({\frac {f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)}{\Delta x}}+{\frac {f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}}\right)\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)}{\Delta x}}+\lim _{\Delta x\to 0}{\frac {f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}g(x+\Delta x)+\lim _{\Delta x\to 0}f(x){\frac {g(x+\Delta x)-g(x)}{\Delta x}}\\&=f'(x)g(x)+f(x)g'(x)&\Box \end{aligned}}$

Kettingregel

$f(x)=g(h(x))\quad geeft\quad f'(x)=g'(h(x))h'(x)$

Bewijs

$(g\circ h)'(x)~=\lim _{x_{0}\rightarrow x}{(g\circ h)(x)-(g\circ h)(x_{0}) \over x-x_{0}}$

$=\lim _{x_{0}\rightarrow x}{g(h(x))-g(h(x_{0})) \over x-x_{0}}$

$=\lim _{x_{0}\rightarrow x}\left[{g(h(x))-g(h(x_{0})) \over h(x)-h(x_{0})}\cdot {h(x)-h(x_{0}) \over x-x_{0}}\right]$

$=\lim _{x_{0}\rightarrow x}{g(h(x))-g(h(x_{0})) \over h(x)-h(x_{0})}\cdot \lim _{x_{0}\rightarrow x}{h(x)-h(x_{0}) \over x-x_{0}}$

$=g'(h(x))\cdot h'(x)$

Quotiëntregel

$f(x)={\frac {g(x)}{h(x)}}\quad geeft\quad f'(x)={\frac {h(x)g'(x)-g(x)h'(x)}{{\Big (}h(x){\Big )}^{2}}}$

Bewijs

$f(x)={\frac {g(x)}{h(x)}}=g(x){\Big (}h(x){\Big )}^{-1}$

Hierop kunnen de product- en de kettingregel worden toegepast:

${\begin{aligned}f'(x)&=g'(x){\Big (}h(x){\Big )}^{-1}+g(x){\bigg [}{\Big (}h(x){\Big )}^{-1}{\bigg ]}'\\&=g'(x){\Big (}h(x){\Big )}^{-1}+g(x){\bigg (}-{\Big (}h(x){\Big )}^{-2}\cdot h'(x){\bigg )}\\&=g'(x){\Big (}h(x){\Big )}^{-1}-g(x)h'(x){\Big (}h(x){\Big )}^{-2}\\&={\frac {h(x)g'(x)}{{\Big (}h(x){\Big )}^{2}}}-{\frac {g(x)h'(x)}{{\Big (}h(x){\Big )}^{2}}}\\&={\frac {h(x)g'(x)-g(x)h'(x)}{{\Big (}h(x){\Big )}^{2}}}&\Box \end{aligned}}$

Afgeleiden Trigonometrische Functies

Sinus

$f(x)=\sin(x)\quad geeft\quad f'(x)=\cos(x)$

Bewijs

${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x)\cos(\Delta x)+\cos(x)\sin({\Delta x})-\sin(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x)\cos(\Delta x)-\sin(x)}{\Delta x}}+\lim _{\Delta x\to 0}{\frac {\cos(x)\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\lim _{\Delta x\to 0}{\frac {cos(\Delta x)-1}{\Delta x}}+\cos(x)\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\lim _{\Delta x\to 0}{\frac {{\Big (}\cos(\Delta x)-1{\Big )}{\Big (}\cos(\Delta x)+1{\Big )}}{\Delta x{\Big (}\cos(\Delta x)+1{\Big )}}}+\cos(x)\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\lim _{\Delta x\to 0}{\frac {{\Big (}\cos ^{2}(\Delta x)-1{\Big )}}{\Delta x{\Big (}\cos(\Delta x)+1{\Big )}}}+\cos(x)\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\lim _{\Delta x\to 0}{\frac {-\sin ^{2}(\Delta x)}{\Delta x{\Big (}\cos(\Delta x)+1{\Big )}}}+\cos(x)\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\lim _{\Delta x\to 0}-\sin(\Delta x)\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\lim _{\Delta x\to 0}{\frac {1}{\cos(\Delta x)+1}}+\cos(x)\lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot 0\cdot 1\cdot {\frac {1}{2}}+\cos(x)\cdot 1\\&=\cos(x)&\Box \end{aligned}}$

Cosinus

$f(x)=\cos(x)\quad geeft\quad f'(x)=-\sin(x)$

Bewijs

Voor dit bewijs maken we gebruik van de kettingregel, de afgeleide van de sinus en de regel dat $\cos(x)=\sin({\tfrac {1}{2}}\pi -x)$.

Substitueer eerst $u={\tfrac {1}{2}}\pi -x$. Dan geldt:

${\begin{aligned}f'(x)&={\frac {{\textrm {d}}f}{{\textrm {d}}u}}\cdot {\frac {{\textrm {d}}u}{{\textrm {d}}x}}\\&=[\sin(u)]'\cdot [u]'\\&=[\sin(u)]'\cdot [{\tfrac {1}{2}}\pi -x]'\\&=\cos(u)\cdot -1\\&=-\cos(u)\\&=-\cos({\tfrac {1}{2}}\pi -x)\\&=-\sin(x)&\Box \end{aligned}}$

Tangens

$f(x)=\tan(x)\quad geeft\quad f'(x)=\sec ^{2}(x)$

Bewijs

We gebruiken bij dit bewijs dat $\tan(x)={\frac {\sin(x)}{\cos(x)}}$.

${\begin{aligned}f'(x)&=[\tan(x)]'\\&={\bigg [}{\frac {\sin(x)}{\cos(x)}}{\bigg ]}'\\&={\frac {\cos(x)\cdot \cos(x)-\sin(x)\cdot -\sin(x)}{cos^{2}(x)}}\\&={\frac {\cos ^{2}(x)+\sin ^{2}(x)}{cos^{2}(x)}}\\&={\frac {1}{cos^{2}(x)}}\\&={\bigg (}{\frac {1}{\cos(x)}}{\bigg )}^{2}\\&=\sec ^{2}(x)&\Box \end{aligned}}$

Secans

$f(x)=\sec(x)\quad geeft\quad f'(x)=\sec(x)\tan(x)$.

Bewijs

We gebruiken bij dit bewijs dat $\sec(x)={\frac {1}{\cos(x)}}$ en dat $\tan(x)={\frac {\sin(x)}{\cos(x)}}$.

${\begin{aligned}f'(x)&=[\sec(x)]'\\&={\bigg [}{\frac {1}{\cos(x)}}{\bigg ]}'\\&={\frac {\cos(x)\cdot 0-1\cdot -\sin(x)}{\cos ^{2}(x)}}\\&={\frac {\sin(x)}{\cos ^{2}(x)}}\\&={\frac {1}{\cos(x)}}\cdot {\frac {\sin(x)}{\cos(x)}}\\&=\sec(x)\tan(x)&\Box \end{aligned}}$

Cosecans

$f(x)=\csc(x)\quad geeft\quad f'(x)=-\cot(x)\csc(x)$

Bewijs

We gebruiken bij dit bewijs dat $\csc(x)={\frac {1}{\sin(x)}}$ en dat $\cot(x)={\frac {1}{\tan(x)}}={\frac {\cos(x)}{\sin(x)}}$. Ook gebruiken we de quotiëntregel.

${\begin{aligned}f'(x)&=[\csc(x)]'\\&={\bigg [}{\frac {1}{\sin(x)}}{\bigg ]}'\\&={\frac {\sin(x)\cdot 0-1\cdot \cos(x)}{\sin ^{2}(x)}}\\&={\frac {-\cos(x)}{\sin ^{2}(x)}}\\&=-{\frac {\cos(x)}{\sin(x)}}\cdot {\frac {1}{\sin(x)}}\\&=-\cot(x)\csc(x)&\Box \end{aligned}}$

Voorbeelden

Somregel:

$f(x)=3x^{3}+2x+4\quad geeft\quad f'(x)=9x^{2}+2$

Productregel:

$c\cdot f(x)\quad geeft\quad {\frac {{\textrm {d}}{\Big (}c\cdot f(x){\Big )}}{{\textrm {d}}x}}=c\cdot f'(x)+c'\cdot f(x)=c\cdot f'(x)$