Question Number 163928 by mathlove last updated on 12/Jan/22
$${f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{100}!} }{\mathrm{100}!}+{x}^{\mathrm{100}} +\mathrm{1} \\ $$$${find}\:\:\:\frac{{d}^{\mathrm{100}!} {f}\left({x}\right)}{{dx}^{\mathrm{100}!} }=? \\ $$
Answered by mr W last updated on 12/Jan/22
$${generally}: \\ $$$${f}\left({x}\right)={x}^{{m}} \\ $$$$\frac{{d}^{{n}} {f}\left({x}\right)}{{dx}^{{n}} }={m}\left({m}−\mathrm{1}\right)\left({m}−\mathrm{2}\right)...\left({m}−{n}+\mathrm{1}\right){x}^{{m}−{n}} \\ $$$$\frac{{d}^{{n}} {f}\left({x}\right)}{{dx}^{{n}} }=\frac{{m}!}{\left({m}−{n}\right)!}{x}^{{m}−{n}} \:\:\:\:{for}\:{n}\leqslant{m} \\ $$$$\frac{{d}^{{n}} {f}\left({x}\right)}{{dx}^{{n}} }=\mathrm{0}\:\:\:\:{for}\:{n}>{m} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{100}!} }{\mathrm{100}!}+{x}^{\mathrm{100}} +\mathrm{1} \\ $$$$\frac{{d}^{\mathrm{100}!} {f}\left({x}\right)}{{dx}^{\mathrm{100}!} }=\frac{\mathrm{2}}{\mathrm{100}!}×\frac{\left(\mathrm{100}!\right)!}{\left(\mathrm{100}!−\mathrm{100}!\right)!}×{x}^{\mathrm{100}!−\mathrm{100}!} +\mathrm{0}+\mathrm{0} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{100}!} {f}\left({x}\right)}{{dx}^{\mathrm{100}!} }=\frac{\mathrm{2}\left(\mathrm{100}!\right)!}{\mathrm{100}!} \\ $$
Commented by mathlove last updated on 12/Jan/22
$${thanks}\:\:{mr} \\ $$