Question Number 145620 by Study last updated on 06/Jul/21
$$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+...}}}\right)=? \\ $$
Answered by Olaf_Thorendsen last updated on 06/Jul/21
$${y}\:=\:{f}\left({x}\right)\:=\:\sqrt[{\mathrm{3}}]{{x}+{f}\left({x}\right)}\:=\:\sqrt[{\mathrm{3}}]{{x}+{y}} \\ $$$${y}^{\mathrm{3}} \:=\:{x}+{y} \\ $$$$\mathrm{3}{y}^{\mathrm{2}} {y}'\:=\:\mathrm{1}+{y}' \\ $$$${y}'\:=\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{3}{y}^{\mathrm{2}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{3}\left({x}+\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+...}}\right)^{\mathrm{2}/\mathrm{3}} −\mathrm{1}} \\ $$