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Differentiëren

naam van de regel	functie	afgeleide
somregel	s\left(x\right)=f\left(x\right)+g\left(x\right)s\left(x\right)=f\left(x\right)+g\left(x\right)s\left(x\right)=f\left(x\right)+g\left(\right)s\left(x\right)=f\left(x\right)+gs\left(x\right)=f\left(x\right)+s\left(x\right)=f\left(x\right)s\left(x\right)=f\left(x\right)s\left(x\right)=f\left(\right)s\left(x\right)=fs\left(x\right)=s\left(x\right)s\left(x\right)s\left(\right)s	s^{\prime}\left(x\right)=f^{\prime}\left(x\right)+g^{\prime}\left(x\right)s\left(x\right)=f^{\prime}\left(x\right)+g^{\prime}\left(x\right)s\left(x\right)=f\left(x\right)+g^{\prime}\left(x\right)s\left(x\right)=f\left(x\right)+g\left(x\right)
verschilregel	v\left(x\right)=f\left(x\right)-g\left(x\right)v\left(x\right)=f\left(x\right)-g\left(x\right)v\left(x\right)=f\left(x\right)-g\left(x\right)v\left(x\right)=f\left(x\right)-g\left(\right)v\left(x\right)=f\left(x\right)-gv\left(x\right)=f\left(x\right)-v\left(x\right)=f\left(x\right)v\left(x\right)=f\left(x\right)v\left(x\right)=f\left(\right)v\left(x\right)=fv\left(x\right)=v\left(x\right)v\left(x\right)v\left(\right)v	v^{\prime}\left(x\right)=f^{\prime}\left(x\right)-g^{\prime}\left(x\right)v^{\prime}\left(x\right)=f^{\prime}\left(x\right)-g\left(x\right)v^{\prime}\left(x\right)=f\left(x\right)-g\left(x\right)v\left(x\right)=f\left(x\right)-g\left(x\right)
productregel	p\left(x\right)=f\left(x\right)\cdot g\left(x\right)p\left(x\right)=f\left(x\right)\cdot g\left(x\right)p\left(x\right)=f\left(x\right)\cdot g\left(\right)p\left(x\right)=f\left(x\right)\cdot gp\left(x\right)=f\left(x\right)\cdotp\left(x\right)=f\left(x\right)p\left(x\right)=f\left(x\right)p\left(x\right)=f\left(\right)p\left(x\right)=fp\left(x\right)=p\left(x\right)p\left(x\right)p\left(\right)p	p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g^{\prime}\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g^{\prime}\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g^{\prime}\left(\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g^{\prime}p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot gp^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdotp^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+f\left(\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+fp^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)+p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot g\left(\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdot gp^{\prime}\left(x\right)=f^{\prime}\left(x\right)\cdotp^{\prime}\left(x\right)=f^{\prime}\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(x\right)p^{\prime}\left(x\right)=f^{\prime}\left(\right)p^{\prime}\left(x\right)=f^{\prime}\left(X\right)p^{\prime}\left(x\right)=f^{\prime}\left(\right)p^{\prime}\left(x\right)=f^{\prime}p^{\prime}\left(x\right)=fp^{\prime}\left(x\right)=p^{\prime}\left(x\right)p^{\prime}\left(x\right)p^{\prime}\left(x_{}\right)p^{\prime}\left(x_{=}\right)p^{\prime}\left(x_{=f}\right)p^{\prime}\left(x_{=f^{}}\right)p^{\prime}\left(x_{=f^{\prime}}\right)p^{\prime}\left(x_{=f^{\prime}\left(\right.}\right)p^{\prime}\left(x_{=f^{\prime}\left(x\right.}\right)p^{\prime}\left(x_{=f^{\prime}\left(x\right)}\right)p^{\prime}\left(x_{=f^{\prime}\left(x\right)}\right)p^{\prime}\left(x_{=f^{\prime}\left(\right)}\right)p^{\prime}\left(x_{=f^{\prime}}\right)p^{\prime}\left(x_{=f}\right)p^{\prime}\left(x_{=}\right)p^{\prime}\left(x\right)p^{\prime}\left(\right)p^{\prime}p
quotiëntregel	q\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}q\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}q\left(x\right)=\frac{f\left(x\right)}{g\left(\right)}q\left(x\right)=\frac{f\left(x\right)}{g}q\left(x\right)=\frac{f\left(x\right)}{\placeholder{}}q\left(x\right)=f\left(x\right)q\left(x\right)=f\left(x\right)q\left(x\right)=f\left(\right)q\left(x\right)=fq\left(x\right)=q\left(x\right)q\left(x\right)q\left(x\right)q\left(\right)qqq\left(\right.q\left(x\right.q\left(x\right)q\left(x\right)_{}q\left(x\right)_{=}q\left(x\right)q\left(x\right)q\left(\right)qqq\left(\right.q\left(x\right.q\left(x\right)q\left(x\right)=q\left(x\right)q\left(x\right)-q\left(x\right)q(xq(x)q(x)\text{=}q(x)\text{=f}q(x)\text{=}	q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(g\left(x\right)\right)^2}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(g\left(x\right)\right)}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(g\left(x\right)\right)}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(g\left(x\right)\right)}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(g\left(\right)\right)}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(g\right)}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\left(\placeholder{}\right)}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}\left(\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g^{\prime}}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f\left(\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-f}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)-}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(g\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(g\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g\left(\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot g}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)\cdot}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(x\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}\left(\right)}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f^{\prime}}{\placeholder{}}q^{\prime}\left(x\right)=\frac{f}{\placeholder{}}q^{\prime}\left(x\right)=\frac{\placeholder{}}{\placeholder{}}q^{\prime}\left(x\right)=q^{\prime}\left(x\right)q^{\prime}\left(x\right)q^{\prime}\left(\right)q^{\prime}q
kettingregel	k\left(x\right)=f\left(g\left(x\right)\right)k\left(x\right)=f\left(g\left(x\right)\right)k\left(x\right)=f\left(g\left(x\right)\right)k\left(x\right)=f\left(g\left(\right)\right)k\left(x\right)=f\left(g\left(c\right)\right)k\left(x\right)=f\left(g\left(\right)\right)k\left(x\right)=f\left(g\right)k\left(x\right)=f\left(\right)k\left(x\right)=f\left(f\right)k\left(x\right)=f\left(\right)k\left(x\right)=fk\left(x\right)=k\left(x\right)k\left(x\right)k\left(\right)k	k^{\prime}\left(x\right)=f^{\prime}\left(g\left(x\right)\right)\cdot g^{\prime}\left(x\right)of\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\cdot\frac{\text{d}g}{\text{d}x}\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\cdot\frac{\text{d}g}{x}\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\cdot\frac{g}{x}\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\cdot\frac{g}{\placeholder{}}\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\cdot g\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\cdot\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{\text{d}g}\frac{\text{d}k}{\text{d}x}=\frac{\text{d}f}{g}\frac{\text{d}k}{\text{d}x}=\frac{f}{g}\frac{\text{d}k}{\text{d}x}=\frac{f}{\placeholder{}}\frac{\text{d}k}{\text{d}x}=f\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}=\frac{\text{d}k}{\text{d}x}\frac{\text{d}k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{x}\frac{k}{\placeholder{}}\frac{\placeholder{}}{\placeholder{}}d

Logaritmen

regel	voorwaarde
^{g}\log\left(a\right)+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+\,^{g}\log\left(\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+\,^{g}\log\left(a\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=^{g}\log\left(ab\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=^{g}\log\left(a\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)=^{g}\log\left(a\right)+^{g}\log\left(a\right)^{g}\log\left(a\right)+^{g}\log\left(a\right)^{g}\log\left(\right)^{g}\log\left(d\right)^{g}\left(d\right)^{g}l\left(d\right)^{g}\left(d\right)^{g}\log_{\placeholder{}}\left(d\right)^{g}\log_{\placeholder{}}\left(\right)^{g}\log_{\placeholder{}}^{g}lo^{g}l^{g}^{g}\log_{\placeholder{}}^{g}lo^{g}l^{g}^{g}\log_{}^{g}\log_{\left.\right)}^{g}\log_{\left(\right)}^{g}\log_{\left(a\right)}^{g}\log_{\left(\placeholder{}\right)}^{g}\log_{\placeholder{}}^{g}lo^{g}l^{g}^{9g}^9	g>0{,}\,g\ne1{,}\,a>0{,}\,b>0g>0{,}\,g\ne1{,}\,a>0{,}\,b>g>0{,}\,g\ne1{,}\,a>0{,}\,bg>0{,}\,g\ne1{,}\,a>0{,}\,g>0{,}\,g\ne1{,}\,a>0{,}g>0{,}\,g\ne1{,}\,a>0{,}g>0{,}\,g\ne1{,}\,a>0{,}g>0{,}\,g\ne1{,}\,a>\frac{0{,}}{}g>0{,}\,g\ne1{,}\,a>\frac{0{,}}{b}g>0{,}\,g\ne1{,}\,a>\frac{0{,}}{\placeholder{}}g>0{,}\,g\ne1{,}\,a>0{,}g>0{,}\,g\ne1{,}\,a>0{,}bg>0{,}\,g\ne1{,}\,a>0{,}g>0{,}\,g\ne1{,}\,a>0g>0{,}\,g\ne1{,}\,a>g>0{,}\,g\ne1{,}\,a>-g>0{,}\,g\ne1{,}\,a>g>0{,}\,g\ne1{,}\,ag>0{,}\,g\ne1{,}\,g>0{,}\,g\ne1{,}g>0{,}\,g\ne1{,}g>0{,}\,g\ne1{,}g>0{,}\,g\ne1g>0{,}\,g\neg>0{,}\,gg>0{,}\,g>0{,}g>0{,}g>0{,}g>0g>g
^{g}\log\left(a\right)-\,^{g}\log\left(b\right)=\,^{g}\log\left(\frac{a}{b}\right)^{g}\log\left(a\right)-\,^{g}\log\left(b\right)=\,^{g}\log\left(\frac{a}{\placeholder{}}\right)^{g}\log\left(a\right)-\,^{g}\log\left(b\right)=\,^{g}\log\left(a\right)^{g}\log\left(a\right)-\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)	g>0{,}\,g\ne1{,}\,a>0{,}\,b>0
^{g}\log\left(a^{p}\right)=p\cdot\,^{g}\log\left(a\right)^{g}\log\left(a^{p}\right)=p\cdot\,^{g}\log\left(a\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot\,^{g}\log\left(\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot\,^{g}\log\left(ba\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot+^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p\cdot+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=p+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)=+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a^{p}\right)+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)	g>0{,}\,g\ne1{,}\,a>0g>0{,}\,g\ne1{,}\,a>0{,}g>0{,}\,g\ne1{,}\,a>0{,}\,g>0{,}\,g\ne1{,}\,a>0{,}\,bg>0{,}\,g\ne1{,}\,a>0{,}\,b>g>0{,}\,g\ne1{,}\,a>0{,}\,b>0
^{g}\log\left(a\right)=\,\frac{^{p}\log\left(a\right)}{^{p}\log\left(g\right)}^{g}\log\left(a\right)=\,\frac{^{p}\log\left(a\right)}{^{p}p\log\left(g\right)}^{g}\log\left(a\right)=\,\frac{^{p}\log\left(a\right)}{p\log\left(g\right)}^{g}\log\left(a\right)=\,\frac{^{p}\log\left(a\right)}{p\log\left(\right)}^{g}\log\left(a\right)=\,\frac{^{p}\log\left(a\right)}{p\log\left(a\right)}^{g}\log\left(a\right)=\,\frac{^{p}\log\left(a\right)}{\placeholder{}}^{g}\log\left(a\right)=\,\frac{\left(^{p}\log\left(a\right)\right.}{\placeholder{}}^{g}\log\left(a\right)=\,\frac{\left(^{p}\log\left(a\right)\right)}{\placeholder{}}^{g}\log\left(a\right)=\,\frac{\left(^{p}\log\left(\right)\right)}{\placeholder{}}^{g}\log\left(a\right)=\,\frac{\left(^{p}\log\left(b\right)\right)}{\placeholder{}}^{g}\log\left(a\right)=\,\frac{\left(^{pg}\log\left(b\right)\right)}{\placeholder{}}^{g}\log\left(a\right)=\,\frac{\left(^{g}\log\left(b\right)\right)}{\placeholder{}}^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)\right)^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{}\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\log\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\log\left(\right.\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\log\left(a\right.\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\log\left(ab\right.\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)\right.^{g}\log\left(a\right)=\,\left(^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)\right)^{g}\log\left(a\right)=\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)\frac{^{g}\log\left(a\right)}{\placeholder{}}\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)^{g}\log\left(a\right)+\,^{g}\log\left(b\right)=\,^{g}\log\left(ab\right)	g>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}\,p\ne1g>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}\,p\neg>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}\,pg>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}\,g>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}g>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}g>0{,}\,g\ne1{,}\,a>0{,}\,p>0{,}g>0{,}\,g\ne1{,}\,a>0{,}\,p>0g>0{,}\,g\ne1{,}\,a>0{,}\,>0g>0{,}\,g\ne1{,}\,a>0{,}\,b>0

2 punten

Open vraag

In figuur 1 zie je een zijaanzicht van drie dominostenen. We nemen in het vervolg van deze opgave het volgende aan:

•De hoogte van de eerste dominosteen is48\operatorname{mm}48\operatorname{m}48\operatorname{cm}48c.

•De dikte van de eerste dominosteen is$7{,}5 \mathrm{~mm}.

•ledere volgende dominosteen is 1,5 keer zo hoog en 1,5 keer zo dik als de voorgaande.

•ledere dominosteen valt tegen het midden van de volgende dominosteen.

•Als de dominostenen op een rij worden gezet, dan krijgt de kleinste steen nummer 1, de volgende steen nummer 2, enzovoorts.

Doordat de eerste dominosteen tegen het midden van de tweede steen moet vallen, ligt de afstandL_1Lvast. Zie figuur 2.

Voor de dominostenen 1 en 2 is deze afstand te berekenen met de formuleL_1=\sqrt{h_{1}{ }^{2}-(0{,}5 \cdot h_{2})^{2}}L_1=\sqrt{h_1^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1\right.^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{}\right.^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,}\right.^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,\,}\right.^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,\,\,}\right.^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,\,\,2}\right.^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,\,\,2}\right)^2-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,\,\,2}\right)-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{\left(h_1^{\,\,\,2}-\left(0{,}5\cdot h_2\right)^2\right.}L_1=\sqrt{\left(h_1^{\,\,\,2}-\left(0{,}5\cdot h_2\right)^2\right)}L_1=\sqrt{\left(h_1^{\,\,\,2}-\left(0{,}5\cdot h_2\right)^2\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}5\cdot h_2\right)^2}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}5\cdot h_2\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}5\cdot h_2\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}5\cdot h\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}5\cdot\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}5\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0{,}\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(0\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(o\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(o,\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(o,5\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(o,\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(o\right)}L_1=\sqrt{h_1^{\,\,\,2}-\left(\right)}L_1=\sqrt{h_1^{\,\,\,2}-}L_1=\sqrt{h_1^{\,\,\,2}-)}L_1=\sqrt{h_1^{\,\,\,2}-}L_1=\sqrt{h_1^{\,\,\,2}}L_1=\sqrt{h_1^{\,\,\,2}}-L_1=\sqrt{h_1^{\,\,\,2}}L_1=\sqrt{h_1^{\,\,2}}L_1=\sqrt{h_1^{\,\,2}}L_1=\sqrt{h_1^{\,\,2}}L_1=\sqrt{h_1^{\,2}}L_1=\sqrt{h_1^{\,2}}L_1=\sqrt{h_1^{\,2}}L_1=\sqrt{h_1^{\,32}}L_1=\sqrt{h_1^{\,32}}L_1=\sqrt{h_1^{\,}}L_1=\sqrt{h_1^{}}L_1=\sqrt{h_1^{}}L_1=\sqrt{h_1^{}}L_1=\sqrt{h_1^{}}L_1=\sqrt{h_1^{}}L_1=\sqrt{h_1^{\&}}L_1=\sqrt{h_1^{}}L_1=\sqrt{h_1^2}L_1=\sqrt{h_1}L_1=\sqrt{h_1w}L_1=\sqrt{h_1}L_1=\sqrt{h}L_1=\sqrt{}L_1=L_1=L_1=L_1=L_1L. Hierbij isL_1L1de afstand tussen dominostenen 1 en 2,h_1hde hoogte van dominosteen 1 enh_2h2de hoogte van dominosteen 2, alle in mm.

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