Question Number 121729 by oustmuchiya@gmail.com last updated on 11/Nov/20
Answered by bemath last updated on 11/Nov/20
$$\left({i}\right)\:\mathrm{2}{yy}'+\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} {y}'=\mathrm{3}{y}' \\ $$$$\Leftrightarrow\:\mathrm{3}{x}^{\mathrm{2}} \:=\:{y}'\left(\mathrm{3}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{2}{y}\right) \\ $$$$\Leftrightarrow\:{y}'\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{3}{y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{3}} \\ $$
Answered by bemath last updated on 11/Nov/20
$$\left({ii}\right)\:{y}'.\mathrm{cos}\:{y}+\mathrm{2}{xy}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} .{y}'+\mathrm{sin}\:{x}=\mathrm{3}{y}' \\ $$$$\Rightarrow\mathrm{2}{xy}^{\mathrm{3}} +\mathrm{sin}\:{x}=\:{y}'\left(\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{cos}\:{y}\right) \\ $$$$\Leftrightarrow\:{y}'\:=\:\frac{\mathrm{2}{xy}^{\mathrm{3}} +\mathrm{sin}\:{x}}{\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{cos}\:{y}}. \\ $$
Answered by bemath last updated on 11/Nov/20
$$\left({iii}\right)\:\mathrm{2}{y}^{\mathrm{3}} −{x}\mathrm{cos}\:{y}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{−\mathrm{1}} {y}^{\mathrm{2}} \:=\:\mathrm{7} \\ $$$$\Rightarrow\mathrm{6}{y}^{\mathrm{2}} .{y}'−\left(\mathrm{cos}\:{y}−{xy}'\mathrm{sin}\:{y}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−{x}^{−\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}{x}^{−\mathrm{1}} {y}.{y}'\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}{y}^{\mathrm{2}} .{y}'−\mathrm{cos}\:{y}+{xy}'\mathrm{sin}\:{y}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }+\frac{{y}.{y}'}{{x}}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{6}{y}^{\mathrm{2}} +{x}\mathrm{sin}\:{y}+\frac{{y}}{{x}}\right).{y}'=\:\frac{{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }+\mathrm{cos}\:{y} \\ $$$$\Rightarrow\left(\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} \mathrm{sin}\:{y}+\mathrm{2}{xy}\right).{y}'={y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \mathrm{cos}\:{y} \\ $$$${y}'=\frac{{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} \mathrm{cos}\:{y}}{\mathrm{12}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} \mathrm{sin}\:{y}+\mathrm{2}{xy}}.\: \\ $$
Answered by bemath last updated on 11/Nov/20
$$\left({iv}\right)\:{using}\:{product}\:{rule} \\ $$$${let}\:\begin{cases}{{u}=\left(\mathrm{5}{x}−\mathrm{3}\right)^{\mathrm{4}} \Rightarrow{u}'=\mathrm{20}\left(\mathrm{5}{x}−\mathrm{3}\right)^{\mathrm{3}} }\\{{v}=\mathrm{3}\left(\mathrm{3}{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \Rightarrow{v}'=\:\mathrm{3}\left(\mathrm{3}{x}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} }\end{cases} \\ $$$${y}'={u}'{v}+{uv}' \\ $$$$ \\ $$