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Question-118064




Question Number 118064 by andilizhaa last updated on 15/Oct/20
Answered by bobhans last updated on 15/Oct/20
f(x)=(((10x^2 −8)^4 ))^(1/5)   by chain rule  letting u = 10x^2 −8 →(du/dx) = 20x  f(x)=(u^4 )^(1/(5 ))  →((df(x))/dx) = ((df(x))/du) × (du/dx)           = (4/(5 (u)^(1/5) )) × (20x) = ((16x)/( ((10x^2 −8))^(1/5) ))
$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{5}}]{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\right)^{\mathrm{4}} } \\ $$$$\mathrm{by}\:\mathrm{chain}\:\mathrm{rule} \\ $$$$\mathrm{letting}\:\mathrm{u}\:=\:\mathrm{10x}^{\mathrm{2}} −\mathrm{8}\:\rightarrow\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{20x} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{5}\:}]{\mathrm{u}^{\mathrm{4}} }\:\rightarrow\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{du}}\:×\:\frac{\mathrm{du}}{\mathrm{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{4}}{\mathrm{5}\:\sqrt[{\mathrm{5}}]{\mathrm{u}}}\:×\:\left(\mathrm{20x}\right)\:=\:\frac{\mathrm{16x}}{\:\sqrt[{\mathrm{5}}]{\mathrm{10x}^{\mathrm{2}} −\mathrm{8}}}\: \\ $$

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