Question Number 202415 by MathematicalUser2357 last updated on 26/Dec/23
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\int{g}'\left({x}\right){f}'\left({g}\left({x}\right)\right){dx}\:\mathrm{is}… \\ $$
Answered by cortano12 last updated on 26/Dec/23
$$\:\mathrm{let}\:\mathrm{u}=\mathrm{g}\left(\mathrm{x}\right)\Rightarrow\mathrm{du}=\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\:\mathrm{I}=\:\int\:\mathrm{g}'\left(\mathrm{x}\right)\:\mathrm{f}\:'\left(\mathrm{g}\left(\mathrm{x}\right)\right)\:\mathrm{dx}\: \\ $$$$\:\:\:=\:\int\:\mathrm{f}\:'\left(\mathrm{u}\right)\:\mathrm{du}=\:\int\:\frac{\mathrm{df}\left(\mathrm{u}\right)}{\mathrm{du}}.\:\mathrm{du} \\ $$$$\:\:=\:\int\:\mathrm{df}\left(\mathrm{u}\right)=\:\mathrm{f}\left(\mathrm{u}\right)+\mathrm{c}\: \\ $$$$\:=\:\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)+\:\mathrm{c}\: \\ $$
Answered by Mathspace last updated on 26/Dec/23
$$=\int\left({fog}\right)^{'} \left({x}\right){dx}={fog}\left({x}\right)+{c} \\ $$
Commented by MathematicalUser2357 last updated on 27/Dec/23
$${f}\circ{g}\:\mathrm{not}\:\mathrm{fog} \\ $$