Question Number 107647 by bobhans last updated on 12/Aug/20
![≍BobHans≍ find the formula (d^3 /dx^3 )[g(f(x))]](https://www.tinkutara.com/question/Q107647.png)
$$\:\:\:\:\:\:\:\asymp\mathcal{B}\mathrm{ob}\mathcal{H}\mathrm{ans}\asymp \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{formula}\:\frac{\mathrm{d}^{\mathrm{3}} }{\mathrm{dx}^{\mathrm{3}} }\left[\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\right] \\ $$
Answered by john santu last updated on 12/Aug/20
![□JS□ (1)(d/dx) g(f(x))= (d/du)g(u) (d/dx)f(x)=g^((1)) f^((1)) =g^((1)) (f(x))f^((1)) (x) (2)(d^2 /dx^2 ) g(f(x))= (d/dx)[(d/dx) g(f(x))] =(d/dx)[(d/du) g(u) (d/dx)f(x)] =[(d/dx) (d/du)g(u)][(d/dx)f(x)]+[(d/du)g(u)][(d^2 /dx^2 )f(x)] = [(du/dx) (d^2 /du^2 )g(u)][(d/dx)f(x)]+[(d/du)g(u)][(d^2 /dx^2 )f(x)] =g^((1)) f^((2)) +g^((2)) [f^((1)) ]^2 (3)(d^3 /dx^3 ) g(f(x))= g^((1)) f^((3)) +3g^((2)) f^((1)) f^((2)) +g^((3)) [f^((1)) ]^3 .](https://www.tinkutara.com/question/Q107651.png)
$$\:\:\:\:\:\square\mathcal{JS}\square \\ $$$$\left(\mathrm{1}\right)\frac{{d}}{{dx}}\:{g}\left({f}\left({x}\right)\right)=\:\frac{{d}}{{du}}{g}\left({u}\right)\:\frac{{d}}{{dx}}{f}\left({x}\right)={g}^{\left(\mathrm{1}\right)} {f}^{\left(\mathrm{1}\right)} \\ $$$$={g}^{\left(\mathrm{1}\right)} \left({f}\left({x}\right)\right){f}^{\left(\mathrm{1}\right)} \left({x}\right) \\ $$$$\left(\mathrm{2}\right)\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\:{g}\left({f}\left({x}\right)\right)=\:\frac{{d}}{{dx}}\left[\frac{{d}}{{dx}}\:{g}\left({f}\left({x}\right)\right)\right] \\ $$$$=\frac{{d}}{{dx}}\left[\frac{{d}}{{du}}\:{g}\left({u}\right)\:\frac{{d}}{{dx}}{f}\left({x}\right)\right] \\ $$$$=\left[\frac{{d}}{{dx}}\:\frac{{d}}{{du}}{g}\left({u}\right)\right]\left[\frac{{d}}{{dx}}{f}\left({x}\right)\right]+\left[\frac{{d}}{{du}}{g}\left({u}\right)\right]\left[\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }{f}\left({x}\right)\right] \\ $$$$=\:\left[\frac{{du}}{{dx}}\:\frac{{d}^{\mathrm{2}} }{{du}^{\mathrm{2}} }{g}\left({u}\right)\right]\left[\frac{{d}}{{dx}}{f}\left({x}\right)\right]+\left[\frac{{d}}{{du}}{g}\left({u}\right)\right]\left[\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }{f}\left({x}\right)\right] \\ $$$$={g}^{\left(\mathrm{1}\right)} {f}^{\left(\mathrm{2}\right)} +{g}^{\left(\mathrm{2}\right)} \left[{f}^{\left(\mathrm{1}\right)} \right]^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\frac{{d}^{\mathrm{3}} }{{dx}^{\mathrm{3}} }\:{g}\left({f}\left({x}\right)\right)=\:{g}^{\left(\mathrm{1}\right)} {f}^{\left(\mathrm{3}\right)} +\mathrm{3}{g}^{\left(\mathrm{2}\right)} {f}^{\left(\mathrm{1}\right)} {f}^{\left(\mathrm{2}\right)} +{g}^{\left(\mathrm{3}\right)} \left[{f}^{\left(\mathrm{1}\right)} \right]^{\mathrm{3}} . \\ $$$$ \\ $$