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Question Number 29831 by abdo imad last updated on 12/Feb/18
$$\left.{let}\:{give}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\right){prove}\:{that}\:\:\:{prove}\:{that} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{{p}_{{n}} \left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:{with}\:{p}_{{n}} {is}\:{a}\:{polynomial} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{p}_{{n}+\mathrm{1}} \left({x}\right)=\left(\mathrm{1}+{x}^{\mathrm{2}} \right){p}_{{n}} ^{'} \left({x}\right)\:−\mathrm{2}\left({n}+\mathrm{1}\right){p}_{{n}} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{p}_{\mathrm{0}} \left({x}\right)\:,{p}_{\mathrm{1}} \left({x}\right)\:,{p}_{\mathrm{2}} \left({x}\right)\:\:,{p}_{\mathrm{3}} \left({x}\right)\:\:. \\ $$