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Question Number 99960 by bemath last updated on 24/Jun/20

$$\mathrm{Given}\mathrm{y}\sqrt{\mathrm{x}}+\mathrm{x}\sqrt{\mathrm{y}}=2.\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}{\mid }_{(1,1)}=?$$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

$$-1$$

Answered by Rio Michael last updated on 24/Jun/20

$$\sqrt{x}\frac{dy}{dx}+\frac{y}{2\sqrt{x}}+\frac{x}{2\sqrt{y}}\frac{dy}{dx}+\sqrt{y}=0$$$$\Rightarrow \sqrt{1}\frac{dy}{dx}{\mid }_{(1,1)}+\frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{1}}\frac{dy}{dx}{\mid }_{(1,1)}+\sqrt{1}=0$$$$$$

Answered by 1549442205 last updated on 24/Jun/20

$$\mathrm{Derivative}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{givenwewe}\mathrm{equation}$$$$\mathrm{by}\mathrm{x}\mathrm{we}\mathrm{obtain}{\mathrm{y}}^{{}^{\prime }}\sqrt{\mathrm{x}}+\frac{\mathrm{y}}{2\sqrt{\mathrm{x}}}+\sqrt{\mathrm{y}}+\frac{{\mathrm{xy}}^{\prime }}{2\sqrt{\mathrm{y}}}=0$$$$\Rightarrow {\mathrm{y}}^{\prime }(\sqrt{\mathrm{x}}+\frac{\mathrm{x}}{2\sqrt{\mathrm{y}}})=-(\frac{\mathrm{y}}{2\sqrt{\mathrm{x}}}+\sqrt{\mathrm{y}})\Rightarrow {\mathrm{y}}^{\prime }=\frac{-(2\sqrt{\mathrm{xy}}+\mathrm{y})}{2\sqrt{\mathrm{xy}}+\mathrm{x}}$$$$\Rightarrow {\mathrm{y}}^{\prime }{\mid }_{(1,1)}=\frac{-(2\sqrt{\mathrm{xy}}+\mathrm{y})}{2\sqrt{\mathrm{xy}}+\mathrm{x}}{\mid }_{(1,1)}=\frac{-3}{3}=-1$$

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