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Question Number 122056 by I want to learn more last updated on 13/Nov/20

Commented by 676597498 last updated on 13/Nov/20

$$\mathrm{your}\mathrm{posting}\mathrm{iymc}\mathrm{questions}$$$$\mathrm{you}\mathrm{will}\mathrm{be}\mathrm{disqualified}$$$$\mathrm{i}\mathrm{have}\mathrm{no}\mathrm{option}\mathrm{but}\mathrm{to}\mathrm{report}$$

Commented by 676597498 last updated on 13/Nov/20

$$\mathrm{This}\mathrm{guy}\mathrm{is}\mathrm{posting}\mathrm{questions}$$$$\mathrm{of}\mathrm{this}\mathrm{competition}(\mathrm{iymc}.\mathrm{info})$$

Answered by mathmax by abdo last updated on 14/Nov/20

$$\mathrm{we}\mathrm{have}{\mathrm{f}}_{\mathrm{m}}\left(\mathrm{x}\right)={\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{e}}^{\mathrm{xlnn}}{\mathrm{x}}^{\mathrm{n}}\right)}^2\Rightarrow$$$${\mathrm{f}}_{\mathrm{m}}^{{}^{\prime }}\left(\mathrm{x}\right)=2\frac{\mathrm{d}}{\mathrm{dx}}\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{xln}\left(\mathrm{n}\right)}\right)\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{xln}\left(\mathrm{n}\right)}$$$$=2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\{{\mathrm{nx}}^{\mathrm{n}-1}{\mathrm{e}}^{\mathrm{xln}\left(\mathrm{n}\right)}+{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{n}\right){\mathrm{n}}^{\mathrm{x}}\}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}{\mathrm{n}}^{\mathrm{x}}$$

Commented by I want to learn more last updated on 14/Nov/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

Commented by mathmax by abdo last updated on 15/Nov/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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