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Question Number 94861 by Ar Brandon last updated on 21/May/20

$$\mathrm{Show}\mathrm{that}$$$${(\mathrm{g}\circ \mathrm{f})}^{\prime }\left(\mathrm{x}\right)={\mathrm{g}}^{\prime }\left(\mathrm{f}\left(\mathrm{x}\right)\right)\cdot {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)$$

Commented by john santu last updated on 21/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{u}\Rightarrow \mathrm{y}=\mathrm{g}\left(\mathrm{u}\right)$$$$\mathrm{chain}\mathrm{rule}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}}\times \frac{\mathrm{du}}{\mathrm{dx}}={\mathrm{g}}^{\prime }\left(\mathrm{u}\right)\times \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)$$$$={\mathrm{g}}^{\prime }\left(\mathrm{f}\left(\mathrm{x}\right)\right)\times \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)$$

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