Question Number 174469 by Mathspace last updated on 01/Aug/22
$${f}\left({x}\right)=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)….\left({x}+{n}\right) \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}^{'} \left({x}\right)\:\:\:\left({n}\geqslant\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right){decompose}\:{F}=\frac{\mathrm{1}}{{f}} \\ $$
Answered by Ar Brandon last updated on 01/Aug/22
$$\mathrm{y}=\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right) \\ $$$$\mathrm{lny}=\mathrm{ln}\left({x}+\mathrm{1}\right)+\mathrm{ln}\left({x}+\mathrm{2}\right)+\centerdot\centerdot\centerdot+\mathrm{ln}\left({x}+{n}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{y}}\centerdot\frac{{d}\mathrm{y}}{{dx}}=\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{{x}+\mathrm{2}}+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{{x}+{n}} \\ $$$$\Rightarrow\frac{{d}\mathrm{y}}{{dx}}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({x}+{k}\right)\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{x}+{k}} \\ $$$$\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)…\left({x}+{n}\right)} \\ $$$$=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!\left({x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({n}−\mathrm{2}\right)!\centerdot\mathrm{1}!\centerdot\left({x}+\mathrm{2}\right)}+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left({n}−\mathrm{1}\right)!\left({x}+{n}\right)} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\left({n}−{k}\right)!\left({k}−\mathrm{1}\right)!\left({x}+{k}\right)} \\ $$