Question Number 59838 by necx1 last updated on 15/May/19
$${What}\:{is}\:{the}\:{nth}\:{derivative}\:{of}\:{sinx}\:{in} \\ $$$${terms}\:{of}\:{the}\:{sine}\:{function}? \\ $$
Commented by maxmathsup by imad last updated on 15/May/19
$${if}\:{f}\left({x}\right)={sinx}\:\:\:,\:\:{f}^{\left({n}\right)} \left({x}\right)\:={sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\:\:\:{with}\:{n}\geqslant\mathrm{1}\:\:{and}\:\:{f}^{\left(\mathrm{0}\right)} ={f} \\ $$$${let}\:{prove}\:{this}\:{by}\:{recurrence}\:{we}\:{have}\:{f}^{\left(\mathrm{1}\right)} \left({x}\right)={cos}\left({x}\right)\:={sin}\left({x}+\frac{\pi}{\mathrm{2}}\right) \\ $$$${let}\:\:{suppose}\:{the}\:{equality}\:{true}\:{at}\:{term}\:{n}\:\:\Rightarrow{f}^{\left({n}+\mathrm{1}\right)} \left({x}\right)\:=\left({f}^{\left({n}\right)} \left({x}\right)\right)^{'} \\ $$$$=\left({sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\right)^{'} \:={cos}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\:={sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\right)\:={sin}\left({x}+\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$${the}\:{result}\:{is}\:{proved}\:. \\ $$
Answered by tanmay last updated on 15/May/19
$${y}={sinx} \\ $$$$\frac{{dy}}{{dx}}={y}_{\mathrm{1}} ={cosx}={sin}\left(\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{2}} =−{sinx}={sin}\left(\mathrm{2}×\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{3}} =−{cosx}={sin}\left(\mathrm{3}×\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{4}} ={sinx}={sin}\left(\mathrm{4}×\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{{n}} ={sin}\left({n}×\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$$${here}\:{n}=\mathrm{9} \\ $$$${y}_{\mathrm{9}} ={sin}\left(\mathrm{9}×\frac{\pi}{\mathrm{2}}+{x}\right) \\ $$
Commented by necx1 last updated on 15/May/19
$${oh}.....\:{I}\:{get}\:{it}\:{now}.\:{Thanks} \\ $$