Analyse/Differentiatie

Afgeleide van een Constante

Voor iedere constante $c$ geldt dat de afgeleide $c^{\prime }=0$:

$f(x)=cgeeft{f}^{\prime }(x)=0$

Bewijs

$\begin{array}{cc}{f}^{\prime }(x)& =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{c-c}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{0}{\mathrm{\Delta }x}\\ & =0& ◻\end{array}$

Afgeleide van een Machtsfunctie

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

$g^{\prime }(x)=na{x}^{n-1}$

Bewijs

Voor gehele, positieve $n$:

Voor dit bewijs gebruiken we het Binomium van Newton:

$\begin{array}{cc}{g}^{\prime }(x)& =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{g(x+\mathrm{\Delta }x)-g(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{a(x+\mathrm{\Delta }x{)}^n-a{x}^n}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{a(\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k(\mathrm{\Delta }x{)}^{n-k})-a{x}^n}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }a\frac{\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k(\mathrm{\Delta }x{)}^{n-k}-x^n}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }a\frac{\sum _{k=0}^{n-1}(\genfrac{}{}{0ex}{}{n}{k})x^k(\mathrm{\Delta }x{)}^{n-k}}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }a(\sum _{k=0}^{n-1}(\genfrac{}{}{0ex}{}{n}{k})x^k(\mathrm{\Delta }x{)}^{n-k-1})\\ & =a(\genfrac{}{}{0ex}{}{n}{n-1})x^{n-1}\\ & =an{x}^{n-1}& ◻\end{array}$

Voor elke $n\in ℝ$:

Nog niet geplaatst.

Afgeleide van een Exponentiële Functie

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

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

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

Bewijs

$\begin{array}{cc}{f}^{\prime }(x)& =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{a^{x+\mathrm{\Delta }x}-a^x}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }{a}^x\frac{a^{\mathrm{\Delta }x}-1}{\mathrm{\Delta }x}\\ & =a^x\cdot \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{a^{\mathrm{\Delta }x}-1}{\mathrm{\Delta }x}\\ & =a^x\cdot \ln (a)& ◻\end{array}$

Somregel

$h(x)=f(x)+g(x)geeft{h}^{\prime }(x)=f^{\prime }(x)+g^{\prime }(x)$

Bewijs

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

$\begin{array}{cc}{h}^{\prime }(x)& =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{h(x+\mathrm{\Delta }x)-h(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{(f(x+\mathrm{\Delta }x)+g(x+\mathrm{\Delta }x))-(f(x)+g(x))}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)+g(x+\mathrm{\Delta }x)-f(x)-g(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)+g(x+\mathrm{\Delta }x)-g(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }(\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}+\frac{g(x+\mathrm{\Delta }x)-g(x)}{\mathrm{\Delta }x})\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}+\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{g(x+\mathrm{\Delta }x)-g(x)}{\mathrm{\Delta }x}\\ & =f^{\prime }(x)+g^{\prime }(x)& ◻\end{array}$

Productregel

$h(x)=f(x)\cdot g(x)geeft{h}^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$

Bewijs

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

$\begin{array}{cc}{h}^{\prime }(x)& =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{h(x+\mathrm{\Delta }x)-h(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)g(x+\mathrm{\Delta }x)-f(x)g(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)g(x+\mathrm{\Delta }x)-f(x)g(x+\mathrm{\Delta }x)+f(x)g(x+\mathrm{\Delta }x)-f(x)g(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }(\frac{f(x+\mathrm{\Delta }x)g(x+\mathrm{\Delta }x)-f(x)g(x+\mathrm{\Delta }x)}{\mathrm{\Delta }x}+\frac{f(x)g(x+\mathrm{\Delta }x)-f(x)g(x)}{\mathrm{\Delta }x})\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)g(x+\mathrm{\Delta }x)-f(x)g(x+\mathrm{\Delta }x)}{\mathrm{\Delta }x}+\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x)g(x+\mathrm{\Delta }x)-f(x)g(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}g(x+\mathrm{\Delta }x)+\underset{\mathrm{\Delta }x\to 0}{\lim }f(x)\frac{g(x+\mathrm{\Delta }x)-g(x)}{\mathrm{\Delta }x}\\ & =f^{\prime }(x)g(x)+f(x)g^{\prime }(x)& ◻\end{array}$

Kettingregel

$f(x)=g(h(x))geeft{f}^{\prime }(x)=g^{\prime }(h(x))h^{\prime }(x)$

Bewijs

$(g\circ h{)}^{\prime }(x)=\underset{x_0\to x}{\lim }\frac{(g\circ h)(x)-(g\circ h)(x_0)}{x-x_0}$

$=\underset{x_0\to x}{\lim }\frac{g(h(x))-g(h(x_0))}{x-x_0}$

$=\underset{x_0\to x}{\lim }[\frac{g(h(x))-g(h(x_0))}{h(x)-h(x_0)}\cdot \frac{h(x)-h(x_0)}{x-x_0}]$

$=\underset{x_0\to x}{\lim }\frac{g(h(x))-g(h(x_0))}{h(x)-h(x_0)}\cdot \underset{x_0\to x}{\lim }\frac{h(x)-h(x_0)}{x-x_0}$

$=g^{\prime }(h(x))\cdot h^{\prime }(x)$

Quotiëntregel

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

Bewijs

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

Hierop kunnen de product- en de kettingregel worden toegepast:

$\begin{array}{cc}{f}^{\prime }(x)& =g^{\prime }(x)(h(x){)}^{-1}+g(x)[(h(x){)}^{-1}{]}^{\prime }\\ & =g^{\prime }(x)(h(x){)}^{-1}+g(x)(-(h(x){)}^{-2}\cdot h^{\prime }(x))\\ & =g^{\prime }(x)(h(x){)}^{-1}-g(x)h^{\prime }(x)(h(x){)}^{-2}\\ & =\frac{h(x)g^{\prime }(x)}{(h(x){)}^2}-\frac{g(x)h^{\prime }(x)}{(h(x){)}^2}\\ & =\frac{h(x)g^{\prime }(x)-g(x)h^{\prime }(x)}{(h(x){)}^2}& ◻\end{array}$

Afgeleiden Trigonometrische Functies

Sinus

$f(x)=\sin (x)geeft{f}^{\prime }(x)=\cos (x)$

Bewijs

$\begin{array}{cc}{f}^{\prime }(x)& =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (x+\mathrm{\Delta }x)-\sin (x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (x)\cos (\mathrm{\Delta }x)+\cos (x)\sin (\mathrm{\Delta }x)-\sin (x)}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (x)\cos (\mathrm{\Delta }x)-\sin (x)}{\mathrm{\Delta }x}+\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\cos (x)\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\\ & =\sin (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{cos(\mathrm{\Delta }x)-1}{\mathrm{\Delta }x}+\cos (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\\ & =\sin (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{(\cos (\mathrm{\Delta }x)-1)(\cos (\mathrm{\Delta }x)+1)}{\mathrm{\Delta }x(\cos (\mathrm{\Delta }x)+1)}+\cos (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\\ & =\sin (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{({\cos }^2(\mathrm{\Delta }x)-1)}{\mathrm{\Delta }x(\cos (\mathrm{\Delta }x)+1)}+\cos (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\\ & =\sin (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{-{\sin }^2(\mathrm{\Delta }x)}{\mathrm{\Delta }x(\cos (\mathrm{\Delta }x)+1)}+\cos (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\\ & =\sin (x)\underset{\mathrm{\Delta }x\to 0}{\lim }-\sin (\mathrm{\Delta }x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{1}{\cos (\mathrm{\Delta }x)+1}+\cos (x)\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\sin (\mathrm{\Delta }x)}{\mathrm{\Delta }x}\\ & =\sin (x)\cdot 0\cdot 1\cdot \frac{1}{2}+\cos (x)\cdot 1\\ & =\cos (x)& ◻\end{array}$

Cosinus

$f(x)=\cos (x)geeft{f}^{\prime }(x)=-\sin (x)$

Bewijs

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

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

$\begin{array}{cc}{f}^{\prime }(x)& =\frac{\text{d}f}{\text{d}u}\cdot \frac{\text{d}u}{\text{d}x}\\ & =[\sin (u){]}^{\prime }\cdot [u{]}^{\prime }\\ & =[\sin (u){]}^{\prime }\cdot [\frac{1}{2}\pi -x{]}^{\prime }\\ & =\cos (u)\cdot -1\\ & =-\cos (u)\\ & =-\cos (\frac{1}{2}\pi -x)\\ & =-\sin (x)& ◻\end{array}$

Tangens

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

Bewijs

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

$\begin{array}{cc}{f}^{\prime }(x)& =[\tan (x){]}^{\prime }\\ & =[\frac{\sin (x)}{\cos (x)}{]}^{\prime }\\ & =\frac{\cos (x)\cdot \cos (x)-\sin (x)\cdot -\sin (x)}{co{s}^2(x)}\\ & =\frac{{\cos }^2(x)+{\sin }^2(x)}{co{s}^2(x)}\\ & =\frac{1}{co{s}^2(x)}\\ & =(\frac{1}{\cos (x)}{)}^2\\ & ={\sec }^2(x)& ◻\end{array}$

Secans

$f(x)=\sec (x)geeft{f}^{\prime }(x)=\sec (x)\tan (x)$.

Bewijs

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

$\begin{array}{cc}{f}^{\prime }(x)& =[\sec (x){]}^{\prime }\\ & =[\frac{1}{\cos (x)}{]}^{\prime }\\ & =\frac{\cos (x)\cdot 0-1\cdot -\sin (x)}{{\cos }^2(x)}\\ & =\frac{\sin (x)}{{\cos }^2(x)}\\ & =\frac{1}{\cos (x)}\cdot \frac{\sin (x)}{\cos (x)}\\ & =\sec (x)\tan (x)& ◻\end{array}$

Cosecans

$f(x)=\csc (x)geeft{f}^{\prime }(x)=-\cot (x)\csc (x)$

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.

$\begin{array}{cc}{f}^{\prime }(x)& =[\csc (x){]}^{\prime }\\ & =[\frac{1}{\sin (x)}{]}^{\prime }\\ & =\frac{\sin (x)\cdot 0-1\cdot \cos (x)}{{\sin }^2(x)}\\ & =\frac{-\cos (x)}{{\sin }^2(x)}\\ & =-\frac{\cos (x)}{\sin (x)}\cdot \frac{1}{\sin (x)}\\ & =-\cot (x)\csc (x)& ◻\end{array}$

Voorbeelden

Somregel:

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

Productregel:

$c\cdot f(x)geeft\frac{\text{d}(c\cdot f(x))}{\text{d}x}=c\cdot f^{\prime }(x)+c^{\prime }\cdot f(x)=c\cdot f^{\prime }(x)$