Question Number 118064 by andilizhaa last updated on 15/Oct/20
Answered by bobhans last updated on 15/Oct/20
$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{5}}]{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\right)^{\mathrm{4}} } \\ $$$$\mathrm{by}\:\mathrm{chain}\:\mathrm{rule} \\ $$$$\mathrm{letting}\:\mathrm{u}\:=\:\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\:\rightarrow\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{20x} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{5}\:}]{\mathrm{u}^{\mathrm{4}} }\:\rightarrow\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{du}}\:×\:\frac{\mathrm{du}}{\mathrm{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{4}}{\mathrm{5}\:\sqrt[{\mathrm{5}}]{\mathrm{u}}}\:×\:\left(\mathrm{20x}\right)\:=\:\frac{\mathrm{16x}}{\:\sqrt[{\mathrm{5}}]{\mathrm{10x}^{\mathrm{2}} −\mathrm{8}}}\: \\ $$