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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

In het dagelijks leven worden er veel batterijen gebruikt om apparaten te laten werken. Een batterij werkt door het verschil in spanning tussen de pluspool en de minpool. Dit spanningsverschil noemen we de totale spanning van de batterij. Deze totale spanning wordt lager als je de batterij enige tijd hebt gebruikt. Hierdoor gaat bijvoorbeeld een fietslamp met een batterij op een gegeven moment minder fel branden.

Een van de eerste batterijen werd in 1836 uitgevonden door de Britse scheikundige Daniell. Bij deze batterij werd bij de pluspool koper en bij de minpool zink gebruikt.

Deze opgave gaat over zo'n batterij.

De spanningen van de polen van deze batterij zijn te berekenen met de volgende formules:

\begin{aligned} & V_{plus}=0{,}34+0{,}0296\cdot\log(k)\\ & V_{min}=-0{,}76+0{,}0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0{,}34+0{,}0296\cdot\log(k)\\ & V_{min}=-0{,}76+00296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0{,}34+0{,}0296\cdot\log(k)\\ & V_{min}=-0{,}76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0{,}34+00296\cdot\log(k)\\ & V_{min}=-0{,}76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0{,}34+0,0296\cdot\log(k)\\ & V_{min}=-0{,}76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0{,}34+0,0296\cdot\log(k)\\ & V_{min}=-076+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0{,}34+0,0296\cdot\log(k)\\ & V_{min}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=034+0,0296\cdot\log(k)\\ & V_{min}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{min}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{main}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mamin}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mamsin}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mamsimn}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mamsim}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mamsi}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mams}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{mam}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{ma}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{m}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}m=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}md=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}m=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mm}}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mm}d}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mm}}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mm}}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mm}d}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mm}}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mi}}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mi}m}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{\operatorname{mi}}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{m}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{n}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{plus}=0,34+0,0296\cdot\log(k)\\ & V_{}=-0,76+0,0296\cdot\log(z)\end{aligned}\begin{aligned} & V_{p l u s}=0,34+0,0296 \cdot \log (k) \\ & V_{\min }=-0,76+0,0296 \cdot \log (z) \end{aligned}

Hierin zijnV_{plus}V_{plu}V_{pl}V_{p}V_{}V_{\text{p}}V_{\text{pl}}V_{\text{plu}}V_{\text{plus}}$V_{\text {plus }}enV_{min}$V_{\text {min }}de spanningen van de polen in volt en$ken$zde concentraties van koper en zink in molair¹⁾.

De totale spanning van de batterij,$V_{\text {batterij }}, is het verschil tussen de spanningen van beide polen:

V_{batterij\text{ }}=V_{plus\text{ }}-V_{min}V_{batteri\text{ }}=V_{plus\text{ }}-V_{min}V_{batter\text{ }}=V_{plus\text{ }}-V_{min}V_{batte\text{ }}=V_{plus\text{ }}-V_{min}V_{batt\text{ }}=V_{plus\text{ }}-V_{min}V_{bat\text{ }}=V_{plus\text{ }}-V_{min}V_{ba\text{ }}=V_{plus\text{ }}-V_{min}V_{b\text{ }}=V_{plus\text{ }}-V_{min}V_{\text{ }}=V_{plus\text{ }}-V_{min}V_{\text{b }}=V_{plus\text{ }}-V_{min}V_{\text{ba }}=V_{plus\text{ }}-V_{min}V_{\text{bat }}=V_{plus\text{ }}-V_{min}V_{\text{batt }}=V_{plus\text{ }}-V_{min}V_{\text{batte }}=V_{plus\text{ }}-V_{min}V_{\text{batter }}=V_{plus\text{ }}-V_{min}V_{\text{batteri }}=V_{plus\text{ }}-V_{min}V_{\text{batterij }}=V_{plus\text{ }}-V_{min}V_{\text{batterij }}=V_{plu\text{ }}-V_{min}V_{\text{batterij }}=V_{pl\text{ }}-V_{min}V_{\text{batterij }}=V_{p\text{ }}-V_{min}V_{\text{batterij }}=V_{\text{ }}-V_{min}V_{\text{batterij }}=V_{\text{p }}-V_{min}V_{\text{batterij }}=V_{\text{pl }}-V_{min}V_{\text{batterij }}=V_{\text{plu }}-V_{min}V_{\text{batterij }}=V_{\text{plus }}-V_{min}V_{\text {batterij }}=V_{\text {plus }}-V_{\min }

noot 1: molair is een maat voor het aantal deeltjes per liter

4 punten

Open vraag

Het is mogelijk om voorV_{batterij}V_{batteri}V_{batter}V_{batte}V_{batt}V_{bat}V_{ba}V_{b}V_{}V_{\text{b}}V_{\text{ba}}V_{\text{bat}}V_{\text{batt}}V_{\text{batte}}V_{\text{batte}}V_{\text{batteri}}V_{\text{batterij}}$V_{\text {batterij }}een formule op te stellen waarin alleen$zvoorkomt:

V_{batterij}=1{,}10+0{,}0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batterij}=1{,}10+00296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batterij}=1{,}10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batterij}=110+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batterij}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batteri}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batter}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batte}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{batt}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{bat}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{ba}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{b}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{b}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{ba}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{bat}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{batt}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{batte}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{batter}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{batteri}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text{batterij}}=1,10+0,0296\cdot\log(2z^{-1}-1),\text{ met }1\leq z<2V_{\text {batterij }}=1,10+0,0296 \cdot \log (2 z^{-1}-1), \text { met } 1 \leq z<2

Doordat$z, de concentratie van het zink, toeneemt tijdens het gebruik van de batterij, daalt de totale spanning van de batterij. Dit kan worden aangetoond door te kijken naar de afgeleide functie vanV_{batterij}V_{batteri}V_{batter}V_{batte}V_{batt}V_{bat}V_{ba}V_{b}V_{}V_{\text{b}}V_{\text{ba}}V_{\text{bat}}V_{\text{batt}}V_{\text{batte}}V_{\text{batter}}V_{\text{batteri}}V_{\text{batteri}}V_{\text{batterij}}$V_{\text {batterij }}. Er geldt (na afronding):

\frac{dV_{batterij\text{ }}}{dz}=\frac{-0{,}0257}{2z-z^2}\frac{dV_{batterij\text{ }}}{dz}=\frac{-00257}{2z-z^2}\frac{dV_{batterij\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{batteri\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{batter\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{batte\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{batt\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{bat\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{ba\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{b\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{ }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{b }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{ba }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{bat }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{bat }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{batte }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{batter }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{dV_{\text{batteri }}}{dz}=\frac{-0,0257}{2 z-z^{2}}\frac{\mathrm{d} V_{\text {batterij }}}{\mathrm{d} z}=\frac{-0,0257}{2 z-z^{2}}

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Bijbehorende onderwerpen

Op deze pagina behandelen we vraag 16 van het centraal examen wiskunde A vwo 2025 – tijdvak 2. Deze vraag is onderdeel van Batterijspanning, en is 4 punten waard.

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Differentiëren 2
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