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Question Number 120288 by Bird last updated on 30/Oct/20
$${let}\:{f}\left({x}\right)=\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$${determine}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$${and}\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Commented by TITA last updated on 30/Oct/20
$${is}\:{it}\:{f}^{'} \left({x}\right)? \\ $$
Answered by TITA last updated on 30/Oct/20
![f^′ (x)=(((x^2 +3)(((2x)/(1+x^2 )))−[(ln (1+x^2 )(2x)])/((x^2 +3)^2 )) f′(x)= (([(x^2 +3)(2x)]−[(1+x^2 )(ln (1+x^2 ))(2x)])/((1+x^2 )(x^2 +3)^2 ))](Q120296.png)
$${f}^{'} \left({x}\right)=\frac{\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)−\left[\left(\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{2}{x}\right)\right]\right.}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)=\:\frac{\left[\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left(\mathrm{2}{x}\right)\right]−\left[\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)\left(\mathrm{2}{x}\right)\right]}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$
Commented by Bird last updated on 30/Oct/20
$${no}\:{f}^{\left({n}\right)} \left({n}^{{eme}} \:{derivstive}!\right) \\ $$