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Question Number 178847 by zaheen last updated on 22/Oct/22

y=xsin (x)  (d^(35) y/dx^(35) )=?

$${y}={x}\mathrm{sin}\:\left({x}\right) \\ $$$$\frac{{d}^{\mathrm{35}} {y}}{{dx}^{\mathrm{35}} }=? \\ $$

Answered by mr W last updated on 22/Oct/22

y=f(x)g(x) with  f(x)=x, g(x)=sin (x)  f^((0) (x)=x,f^((1)) (x)=1, f^((k)) (x)=0 for k≥2  g^((0) (x)=sin x, g^((1)) (x)=cos x, g^((2)) (x)=−sin x  g^((2k+1)) (x)=(−1)^k cos x, g^((2k)) (x)=(−1)^k sin x  y^((35)) =f^((0)) (x)g^((35)) (x)+35f^((1)) (x)g^((34)) (x)+0          =x(−cos x)+35(1)(−sin x)          =−x cos x−35 sin x

$${y}={f}\left({x}\right){g}\left({x}\right)\:{with} \\ $$$${f}\left({x}\right)={x},\:{g}\left({x}\right)=\mathrm{sin}\:\left({x}\right) \\ $$$${f}^{\left(\mathrm{0}\right.} \left({x}\right)={x},{f}^{\left(\mathrm{1}\right)} \left({x}\right)=\mathrm{1},\:{f}^{\left({k}\right)} \left({x}\right)=\mathrm{0}\:{for}\:{k}\geqslant\mathrm{2} \\ $$$${g}^{\left(\mathrm{0}\right.} \left({x}\right)=\mathrm{sin}\:{x},\:{g}^{\left(\mathrm{1}\right)} \left({x}\right)=\mathrm{cos}\:{x},\:{g}^{\left(\mathrm{2}\right)} \left({x}\right)=−\mathrm{sin}\:{x} \\ $$$${g}^{\left(\mathrm{2}{k}+\mathrm{1}\right)} \left({x}\right)=\left(−\mathrm{1}\right)^{{k}} \mathrm{cos}\:{x},\:{g}^{\left(\mathrm{2}{k}\right)} \left({x}\right)=\left(−\mathrm{1}\right)^{{k}} \mathrm{sin}\:{x} \\ $$$${y}^{\left(\mathrm{35}\right)} ={f}^{\left(\mathrm{0}\right)} \left({x}\right){g}^{\left(\mathrm{35}\right)} \left({x}\right)+\mathrm{35}{f}^{\left(\mathrm{1}\right)} \left({x}\right){g}^{\left(\mathrm{34}\right)} \left({x}\right)+\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:={x}\left(−\mathrm{cos}\:{x}\right)+\mathrm{35}\left(\mathrm{1}\right)\left(−\mathrm{sin}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:=−{x}\:\mathrm{cos}\:{x}−\mathrm{35}\:\mathrm{sin}\:{x} \\ $$

Commented by Tawa11 last updated on 23/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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