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Question Number 99357 by bobhans last updated on 20/Jun/20

$${\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2-3\mathrm{y}=10,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{at}\mathrm{x}=2$$

Commented by bemath last updated on 20/Jun/20

$$\mathrm{x}=2\Rightarrow 4+2\mathrm{y}+{\mathrm{y}}^2-3\mathrm{y}-10=0$$$${\mathrm{y}}^2-\mathrm{y}-6=0\Rightarrow \{\begin{array}{l}\mathrm{y}=3\\ \mathrm{y}=-2\end{array}$$$$\mathrm{implicit}\mathrm{differentiate}$$$$2\mathrm{x}+\mathrm{y}+\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}-3\frac{\mathrm{dy}}{\mathrm{dx}}=0$$$$2\mathrm{x}+\mathrm{y}+(\mathrm{x}-3+2\mathrm{y})\frac{\mathrm{dy}}{\mathrm{dx}}=0$$$$\mathrm{for}(2,3)\Rightarrow 4+3+(2-3+6)\frac{\mathrm{dy}}{\mathrm{dx}}=0$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{7}{5}$$$$\mathrm{for}(2,-2)\Rightarrow 4-2+(2-3-4)\frac{\mathrm{dy}}{\mathrm{dx}}=0$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{(-5)}=\frac{2}{5}$$

Answered by mathmax by abdo last updated on 20/Jun/20

$${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}-3\mathrm{y}=10\Rightarrow {\mathrm{y}}^2+(\mathrm{x}-3)\mathrm{y}+{\mathrm{x}}^2-10=0$$$$\mathrm{\Delta }={(\mathrm{x}-3)}^2-4({\mathrm{x}}^2-10)={\mathrm{x}}^2-6\mathrm{x}+9-{4\mathrm{x}}^2+40=-{3\mathrm{x}}^2-6\mathrm{x}+49$$$$\mathrm{for}\mathrm{\Delta }⩾0\mathrm{we}\mathrm{get}{\mathrm{y}}_1=\frac{3-\mathrm{x}+\sqrt{-{3\mathrm{x}}^2-6\mathrm{x}+49}}{2}$$$$\mathrm{and}{\mathrm{y}}_2=\frac{3-\mathrm{x}-\sqrt{-{3\mathrm{x}}^2-6\mathrm{x}+49}}{2}$$$$\mathrm{y}={\mathrm{y}}_1\Rightarrow {\mathrm{y}}^{{}^{\prime }}=-\frac{1}{2}+\frac{1}{2}\times \frac{-6\mathrm{x}-6}{2\sqrt{-{3\mathrm{x}}^2-6\mathrm{x}+49}}=-\frac{1}{2}-\frac{3\mathrm{x}+3}{2\sqrt{-{3\mathrm{x}}^2-6\mathrm{x}+49}}$$$${\mathrm{y}}^{{}^{\prime }}\left(2\right)=-\frac{1}{2}-\frac{9}{2\sqrt{-12-12+49}}=-\frac{1}{2}-\frac{9}{10}=\frac{-5-9}{10}=\frac{-14}{10}=-\frac{7}{5}$$$$\mathrm{y}={\mathrm{y}}_2\Rightarrow {\mathrm{y}}^{{}^{\prime }}=-\frac{1}{2}-\frac{1}{2}\times \frac{-6\mathrm{x}-6}{2\sqrt{-{3\mathrm{x}}^2-6\mathrm{x}+49}}=-\frac{1}{2}+\frac{1}{2}\frac{3\mathrm{x}+3}{\sqrt{-{3\mathrm{x}}^2-6\mathrm{x}+49}}\Rightarrow$$$${\mathrm{y}}^{{}^{\prime }}\left(2\right)=-\frac{1}{2}+\frac{9}{2\times 5}=-\frac{1}{2}+\frac{9}{10}=\frac{-5+9}{10}=\frac{4}{10}=\frac{2}{5}$$

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