Question Number 201940 by Calculusboy last updated on 15/Dec/23
$$\boldsymbol{{tan}}^{\mathrm{3}} \left(\boldsymbol{{xy}}^{\mathrm{2}} +\boldsymbol{{y}}\right)=\boldsymbol{{x}}\:\:\boldsymbol{{find}}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}} \\ $$
Answered by cortano12 last updated on 16/Dec/23
![⇒(d/dx) [ tan^3 (xy^2 +y) ] = (d/dx)(x) ⇒ 3tan^2 (xy^2 +y) sec^2 (xy^2 +y) (y^2 +2xyy^′ +y^′ )=1 ⇒ y^2 +(2xy+1)y^′ = ((cos^2 (xy^2 +y))/(3tan^2 (xy^2 +y))) ⇒y^′ = (1/(2xy+1)) (((cos^2 (xy^2 +y))/(3tan^2 (xy^2 +y))) −y^2 )](https://www.tinkutara.com/question/Q201945.png)
$$\:\:\Rightarrow\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:\mathrm{tan}\:^{\mathrm{3}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)\:\right]\:=\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right) \\ $$$$\:\Rightarrow\:\mathrm{3tan}\:^{\mathrm{2}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)\:\mathrm{sec}\:^{\mathrm{2}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)\:\left(\mathrm{y}^{\mathrm{2}} +\mathrm{2xyy}^{'} +\mathrm{y}^{'} \right)=\mathrm{1} \\ $$$$\:\Rightarrow\:\mathrm{y}^{\mathrm{2}} +\left(\mathrm{2xy}+\mathrm{1}\right)\mathrm{y}^{'} =\:\frac{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)}{\mathrm{3tan}\:^{\mathrm{2}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)} \\ $$$$\:\Rightarrow\mathrm{y}^{'} =\:\frac{\mathrm{1}}{\mathrm{2xy}+\mathrm{1}}\:\left(\frac{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)}{\mathrm{3tan}\:^{\mathrm{2}} \left(\mathrm{xy}^{\mathrm{2}} +\mathrm{y}\right)}\:−\mathrm{y}^{\mathrm{2}} \right)\: \\ $$
Commented by Calculusboy last updated on 16/Dec/23
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{sir}} \\ $$