uitgewerkt

$
\eqalign{
&f(x)=\frac{1}
{{x^2}}\cr
&f(x)=x^{-2}\cr
&f'(x)=-2\cdot x^{-3}\cr
&f'(x)=-\frac{2}
{{x^3}}\cr}
$
$
\eqalign{
&g(x)=\sqrt x\cr
&g(x)=x^{\frac{1}
{2}}\cr
&g'(x)=\frac{1}
{2}x^{-\frac{1}
{2}}\cr
&g'(x)=\frac{1}
{2}\cdot\frac{1}
{{x^{\frac{1}
{2}}}}\cr
&g'(x)=\frac{1}
{2}\cdot\frac{1}
{{\sqrt x}}\cr
&g'(x)=\frac{1}
{{2\sqrt x}}\cr}
$
$
\eqalign{
&h(x)=\frac{2}
{{\root3\of{x^2}}}\cr
&h(x)=\frac{2}
{{x^{\frac{2}
{3}}}}\cr
&h(x)=2\cdot x^{-\frac{2}
{3}}\cr
&h'(x)=2\cdot-\frac{2}
{3}x^{-1\frac{2}
{3}}\cr
&h'(x)=-\frac{4}
{3}x^{-\frac{5}
{3}}\cr
&h'(x)=-\frac{4}
{3}\cdot\frac{1}
{{x^{\frac{5}
{3}}}}\cr
&h'(x)=-\frac{4}
{3}\cdot\frac{1}
{{\root3\of{x^5}}}\cr
&h'(x)=-\frac{4}
{{3\root3\of{x^5}}}\cr}
$

$
\eqalign{
&k(x)=\frac{{x^3-4}}
{{3x}}\cr
&k(x)=\frac{{x^3}}
{{3x}}-\frac{4}
{{3x}}\cr
&k(x)=\frac{1}
{3}x^2-\frac{4}
{3}x^{-1}\cr
&k'(x)=\frac{1}
{3}\cdot2x-\frac{4}
{3}\cdot-1\cdot x^{-2}\cr
&k'(x)=\frac{2}
{3}x+\frac{4}
{3}\cdot\frac{1}
{{x{}^2}}\cr
&k'(x)=\frac{2}
{3}x+\frac{4}
{{3x{}^2}}\cr}$
$\eqalign{
&k'(x)=\frac{2}
{3}x\cdot\frac{{x^2}}
{{x^2}}+\frac{4}
{{3x^2}}\cr
&k'(x)=\frac{{2x^3}}
{{3x^2}}+\frac{4}
{{3x^2}}\cr
&k'(x)=\frac{{2x^3+4}}
{{3x^2}}\cr}
$

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