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Question Number 201940 by Calculusboy last updated on 15/Dec/23

$${\mathit{t}\mathit{a}\mathit{n}}^3({\mathit{x}\mathit{y}}^2+\mathit{y})=\mathit{x}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$$

Answered by cortano12 last updated on 16/Dec/23

$$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\tan {}^3({\mathrm{xy}}^2+\mathrm{y})\right]=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right)$$$$\Rightarrow 3\tan {}^2({\mathrm{xy}}^2+\mathrm{y})\sec {}^2({\mathrm{xy}}^2+\mathrm{y})({\mathrm{y}}^2+{2\mathrm{xyy}}^{{}^{\prime }}+{\mathrm{y}}^{{}^{\prime }})=1$$$$\Rightarrow {\mathrm{y}}^2+(2\mathrm{xy}+1){\mathrm{y}}^{{}^{\prime }}=\frac{\cos {}^2({\mathrm{xy}}^2+\mathrm{y})}{3\tan {}^2({\mathrm{xy}}^2+\mathrm{y})}$$$$\Rightarrow {\mathrm{y}}^{{}^{\prime }}=\frac{1}{2\mathrm{xy}+1}(\frac{\cos {}^2({\mathrm{xy}}^2+\mathrm{y})}{3\tan {}^2({\mathrm{xy}}^2+\mathrm{y})}-{\mathrm{y}}^2)$$

Commented by Calculusboy last updated on 16/Dec/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{r}$$

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