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

let f(x)=x^7 arctan(2x) 1)calculate f^((4)) (0) and f^((7)) (0) 2) calculate f^((5)) (1)

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

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

f(x)=x^7 arctan(2x) ⇒f^((n)) (x)=Σ_(k=0) ^n C_n ^k (x^7 )^((k)) (arctan(2x))^((n−k)) =x^7 (arctan(2x))^((n)) +C_n ^1 7x^6 (arctan(2x))^((n−1)) +7.6C_n ^2 x^5 (arctan(2x))^((n−2)) +7.6.5 C_n ^3 x^4 (arctan(2x))^((n−3)) +7.6.5.4 C_n ^4 x^3 (arctan(2x))^((n−4)) +7.6.5.4.3 C_n ^5 x^2 (arctan(2x))^((n−5)) +7.6.5.4.3.2C_n ^6 x(arctan(2x))^((n−6)) +7.6.5.4.3.2.1 C_n ^7 (arctan(2x))^((n−7)) we have (arctan(2x))^((1)) =(2/(1+4x^2 )) =(2/(4(x^2 +(1/4)))) =(1/(2(x−(i/2))(x+(i/2)))) =(1/(2×((2i)/2)))((1/(x−(i/2)))−(1/(x+(i/2))))=(1/(2i))((1/(x−(i/2)))−(1/(x+(i/2)))) ⇒ (arctan(2x)^((m)) =(1/(2i)){ (((−1)^(m−1) (m−1)!)/((x−(i/2))^m ))−(((−1)^(m−1) (m−1)!)/((x+(i/2))^m ))} =(((−1)^(m−1) (m−1)!)/(2i)){((2iIm(x+(i/2))^m )/((x^2 +(1/4))^m ))} =(−1)^(m−1) (m−1)!×((Im(x+(i/2))^m )/((x^2 +(1/4))^m ))....be continued...

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{7}} \:\mathrm{arctan}\left(\mathrm{2x}\right)\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \left(\mathrm{x}^{\mathrm{7}} \right)^{\left(\mathrm{k}\right)} \:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\mathrm{x}^{\mathrm{7}} \:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}\right)} \:+\mathrm{C}_{\mathrm{n}} ^{\mathrm{1}} \mathrm{7x}^{\mathrm{6}} \left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{1}\right)} \:+\mathrm{7}.\mathrm{6C}_{\mathrm{n}} ^{\mathrm{2}} \mathrm{x}^{\mathrm{5}} \left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{2}\right)} \\ $$$$+\mathrm{7}.\mathrm{6}.\mathrm{5}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{3}} \:\mathrm{x}^{\mathrm{4}} \left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{3}\right)} \:+\mathrm{7}.\mathrm{6}.\mathrm{5}.\mathrm{4}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{4}} \:\mathrm{x}^{\mathrm{3}} \left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{4}\right)} \\ $$$$+\mathrm{7}.\mathrm{6}.\mathrm{5}.\mathrm{4}.\mathrm{3}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{5}} \:\mathrm{x}^{\mathrm{2}} \left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{5}\right)} \:+\mathrm{7}.\mathrm{6}.\mathrm{5}.\mathrm{4}.\mathrm{3}.\mathrm{2C}_{\mathrm{n}} ^{\mathrm{6}} \mathrm{x}\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{6}\right)} \\ $$$$+\mathrm{7}.\mathrm{6}.\mathrm{5}.\mathrm{4}.\mathrm{3}.\mathrm{2}.\mathrm{1}\:\mathrm{C}_{\mathrm{n}} ^{\mathrm{7}} \:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{n}−\mathrm{7}\right)} \\ $$$$\mathrm{we}\:\mathrm{have}\:\left(\mathrm{arctan}\left(\mathrm{2x}\right)\right)^{\left(\mathrm{1}\right)} \:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4x}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}\right)\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{2i}}{\mathrm{2}}}\left(\frac{\mathrm{1}}{\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\mathrm{2i}}\left(\frac{\mathrm{1}}{\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\left(\mathrm{arctan}\left(\mathrm{2x}\right)^{\left(\mathrm{m}\right)} \:=\frac{\mathrm{1}}{\mathrm{2i}}\left\{\:\frac{\left(−\mathrm{1}\right)^{\mathrm{m}−\mathrm{1}} \left(\mathrm{m}−\mathrm{1}\right)!}{\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{m}} }−\frac{\left(−\mathrm{1}\right)^{\mathrm{m}−\mathrm{1}} \left(\mathrm{m}−\mathrm{1}\right)!}{\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{m}} }\right\}\right. \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{m}−\mathrm{1}} \left(\mathrm{m}−\mathrm{1}\right)!}{\mathrm{2i}}\left\{\frac{\mathrm{2iIm}\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{m}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{m}} }\right\} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{m}−\mathrm{1}} \left(\mathrm{m}−\mathrm{1}\right)!×\frac{\mathrm{Im}\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{2}}\right)^{\mathrm{m}} }{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{m}} }....\mathrm{be}\:\mathrm{continued}... \\ $$

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