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Question Number 130534 by mathmax by abdo last updated on 26/Jan/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^7\arctan \left(2\mathrm{x}\right)$$$$1)\mathrm{calculate}{\mathrm{f}}^{\left(4\right)}\left(0\right)\mathrm{and}{\mathrm{f}}^{\left(7\right)}\left(0\right)$$$$2)\mathrm{calculate}{\mathrm{f}}^{\left(5\right)}\left(1\right)$$

Answered by mathmax by abdo last updated on 30/Jan/21

$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^7\arctan \left(2\mathrm{x}\right)\Rightarrow {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\left({\mathrm{x}}^7\right)}^{\left(\mathrm{k}\right)}{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-\mathrm{k})}$$$$={\mathrm{x}}^7{\left(\arctan \left(2\mathrm{x}\right)\right)}^{\left(\mathrm{n}\right)}+{\mathrm{C}}_{\mathrm{n}}^1{7\mathrm{x}}^6{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-1)}+7.{6\mathrm{C}}_{\mathrm{n}}^2{\mathrm{x}}^5{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-2)}$$$$+7.6.5{\mathrm{C}}_{\mathrm{n}}^3{\mathrm{x}}^4{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-3)}+7.6.5.4{\mathrm{C}}_{\mathrm{n}}^4{\mathrm{x}}^3{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-4)}$$$$+7.6.5.4.3{\mathrm{C}}_{\mathrm{n}}^5{\mathrm{x}}^2{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-5)}+7.6.5.4.3.{2\mathrm{C}}_{\mathrm{n}}^6\mathrm{x}{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-6)}$$$$+7.6.5.4.3.2.1{\mathrm{C}}_{\mathrm{n}}^7{\left(\arctan \left(2\mathrm{x}\right)\right)}^{(\mathrm{n}-7)}$$$$\mathrm{we}\mathrm{have}{\left(\arctan \left(2\mathrm{x}\right)\right)}^{\left(1\right)}=\frac{2}{1+{4\mathrm{x}}^2}=\frac{2}{4({\mathrm{x}}^2+\frac{1}{4})}$$$$=\frac{1}{2(\mathrm{x}-\frac{\mathrm{i}}{2})(\mathrm{x}+\frac{\mathrm{i}}{2})}=\frac{1}{2\times \frac{2\mathrm{i}}{2}}(\frac{1}{\mathrm{x}-\frac{\mathrm{i}}{2}}-\frac{1}{\mathrm{x}+\frac{\mathrm{i}}{2}})=\frac{1}{2\mathrm{i}}(\frac{1}{\mathrm{x}-\frac{\mathrm{i}}{2}}-\frac{1}{\mathrm{x}+\frac{\mathrm{i}}{2}})\Rightarrow$$$$(\arctan {\left(2\mathrm{x}\right)}^{\left(\mathrm{m}\right)}=\frac{1}{2\mathrm{i}}\{\frac{{(-1)}^{\mathrm{m}-1}(\mathrm{m}-1)!}{{(\mathrm{x}-\frac{\mathrm{i}}{2})}^{\mathrm{m}}}-\frac{{(-1)}^{\mathrm{m}-1}(\mathrm{m}-1)!}{{(\mathrm{x}+\frac{\mathrm{i}}{2})}^{\mathrm{m}}}\}$$$$=\frac{{(-1)}^{\mathrm{m}-1}(\mathrm{m}-1)!}{2\mathrm{i}}\left\{\frac{2\mathrm{iIm}{(\mathrm{x}+\frac{\mathrm{i}}{2})}^{\mathrm{m}}}{{({\mathrm{x}}^2+\frac{1}{4})}^{\mathrm{m}}}\right\}$$$$={(-1)}^{\mathrm{m}-1}(\mathrm{m}-1)!\times \frac{\mathrm{Im}{(\mathrm{x}+\frac{\mathrm{i}}{2})}^{\mathrm{m}}}{{({\mathrm{x}}^2+\frac{1}{4})}^{\mathrm{m}}}....\mathrm{be}\mathrm{continued}...$$

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