Question Number 132104 by mnjuly1970 last updated on 11/Feb/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{calculus}\:\left(\mathrm{1}\right)… \\ $$$$\:\:{if}\:\:{y}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{sin}\left({x}\right)+{cos}\left({x}\right)}}\:\:{then}\:: \\ $$$$\:\:\:\:\mathrm{3}{y}''−\mathrm{12}\left({y}'\right)^{\mathrm{2}} −{y}^{\mathrm{2}} =\:??? \\ $$$$\:\:\:\:\:……… \\ $$
Answered by mnjuly1970 last updated on 11/Feb/21
$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\frac{\mathrm{1}}{{y}^{\mathrm{3}} }={sin}\left({x}\right)+{cos}\left({x}\right) \\ $$$$\:{differentiation}\:{both}\:{sides}:\:\:\:\frac{−\mathrm{3}{y}'{y}^{\mathrm{2}} }{{y}^{\mathrm{6}} }={cos}\left({x}\right)−{sin}\left({x}\right) \\ $$$$\:\:\:\:\:{simplification}::\:\:\frac{\mathrm{3}{y}'}{{y}^{\mathrm{4}} }={sin}\left({x}\right)−{cos}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:{differentiation}\:{again}::\:\frac{\mathrm{3}{y}''{y}^{\mathrm{4}} −\mathrm{12}\left({y}'\right)^{\mathrm{2}} {y}^{\mathrm{3}} }{{y}^{\mathrm{8}} }\:={cos}\left({x}\right)+{sin}\left({x}\right)=\frac{\mathrm{1}}{{y}^{\mathrm{3}} }\:\:\:\: \\ $$$${simplification}::\:\frac{\mathrm{3}{y}''{y}−\mathrm{12}\left({y}'\right)^{\mathrm{2}} }{{y}^{\mathrm{5}} }=\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\therefore\:\:\mathrm{3}{y}''{y}−\mathrm{12}\left({y}'\right)^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\: \\ $$