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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

3 punten

Open vraag

UitV_{batterij}=V_{plus}-V_{min}V_{batterij}=V_{plus}-V_{mcin}V_{batterij}=V_{plus}-V_{mcicn}V_{batterij}=V_{plus}-V_{mciccn}V_{batterij}=V_{plus}-V_{mcicc}V_{batterij}=V_{plus}-V_{mcic}V_{batterij}=V_{plus}-V_{mcic\in}V_{batterij}=V_{plus}-V_{mcic}V_{batterij}=V_{plus}-V_{mcic\in}V_{batterij}=V_{plus}-V_{mcici}V_{batterij}=V_{plus}-V_{mcic}V_{batterij}=V_{plus}-V_{mci}V_{batterij}=V_{plus}-V_{mc}V_{batterij}=V_{plus}-V_{m}V_{batterij}=V_{plus}-V_{}V_{batterij}=V_{plus}-V_{\text{ }}V_{batterij}=V_{plus}-V_{\text{m }}V_{batterij}=V_{plus}-V_{\text{mi }}V_{batterij}=V_{plus}-V_{\text{min }}V_{batterij}=V_{plu}-V_{\text{min }}V_{batterij}=V_{pl}-V_{\text{min }}V_{batterij}=V_{p}-V_{\text{min }}V_{batterij}=V_{}-V_{\text{min }}V_{batterij}=V_{\text{p}}-V_{\text{min }}V_{batterij}=V_{\text{pl}}-V_{\text{min }}V_{batterij}=V_{\text{plu}}-V_{\text{min }}V_{batterij}=V_{\text{plus}}-V_{\text{min }}V_{batterij}=V_{\text{plus }}-V_{\text{min }}V_{batterij\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{batteri\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{batter\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{batte\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{batt\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{bat\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{ba\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{b\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{\text{ }}=V_{\text{plus }}-V_{\text{min }}V_{\text{b }}=V_{\text{plus }}-V_{\text{min }}V_{\text{ba }}=V_{\text{plus }}-V_{\text{min }}V_{\text{bat }}=V_{\text{plus }}-V_{\text{min }}V_{\text{batt }}=V_{\text{plus }}-V_{\text{min }}V_{\text{batte }}=V_{\text{plus }}-V_{\text{min }}V_{\text{batter }}=V_{\text{plus }}-V_{\text{min }}V_{\text{batteri }}=V_{\text{plus }}-V_{\text{min }}$V_{\text {batterij }}=V_{\text {plus }}-V_{\text {min }}en de gegeven formules voorV_{plus}V_{plu}V_{pl}V_{p}V_{}V_{\text{p}}V_{\text{pl}}V_{\text{plu}}V_{\text{plus}}$V_{\text {plus }}enV_{min}V_{mdin}V_{mdidn}V_{mdid}V_{mdi}V_{md}V_{m}V_{}V_{\min}V_{\operatorname{mi}}V_{m}V_{}V_{\text{m}}V_{\text{mi}}V_{\text{min}}$V_{\text {min }}kan de volgende formule worden herleid:

V_{batterij}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{batteri}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{batter}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{batte}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{batt}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{bat}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{ba}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{b}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{b}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{ba}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{bat}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batt}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batte}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batter}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batteri}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batterij}}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batterij }}=1{,}10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batterij }}=110+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batterij }}=1,10+0{,}0296\cdot\log(\frac{k}{z})V_{\text{batterij }}=1,10+00296\cdot\log(\frac{k}{z})V_{\text {batterij }}=1,10+0,0296 \cdot \log (\frac{k}{z})

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Op deze pagina behandelen we vraag 15 van het centraal examen wiskunde A vwo 2025 – tijdvak 2. Deze vraag is onderdeel van Batterijspanning, en is 3 punten waard.

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