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Question Number 118064 by andilizhaa last updated on 15/Oct/20

Answered by bobhans last updated on 15/Oct/20

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[5]{{({10\mathrm{x}}^2-8)}^4}$$$$\mathrm{by}\mathrm{chain}\mathrm{rule}$$$$\mathrm{letting}\mathrm{u}={10\mathrm{x}}^2-8\to \frac{\mathrm{du}}{\mathrm{dx}}=20\mathrm{x}$$$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt[5]{{\mathrm{u}}^4}\to \frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{du}}\times \frac{\mathrm{du}}{\mathrm{dx}}$$$$=\frac{4}{5\sqrt[5]{\mathrm{u}}}\times \left(20\mathrm{x}\right)=\frac{16\mathrm{x}}{\sqrt[5]{{10\mathrm{x}}^2-8}}$$

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