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Question Number 97073 by naka3546 last updated on 06/Jun/20

$$Find\frac{dy}{dx}of{2}^x+y^2=2xy,x,y\in \mathbb{C}$$

Commented by john santu last updated on 06/Jun/20

$$2^x\ln \left(2\right)+{2\mathrm{yy}}^{\prime }=2\mathrm{y}+2x{\mathrm{y}}^{\prime }$$$${2\mathrm{yy}}^{\prime }-2x{\mathrm{y}}^{\prime }=2\mathrm{y}-2^x\ln \left(2\right)$$$${\mathrm{y}}^{\prime }(2\mathrm{y}-2x)=2\mathrm{y}-2^x\ln \left(2\right)$$$${\mathrm{y}}^{\prime }=\frac{\mathrm{y}-2^{x-1}\ln \left(2\right)}{(\mathrm{y}-x)}$$

Commented by mahdi last updated on 06/Jun/20

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(2^{\mathrm{x}}\right)=2^{\mathrm{x}}\times \ln 2\mathrm{not}\left(\frac{2^{\mathrm{x}}}{\ln 2}\right)$$

Commented by john santu last updated on 06/Jun/20

$$\mathrm{oo}\mathrm{sorry}$$

Commented by Rasheed.Sindhi last updated on 06/Jun/20

$$PlMrnaka,confirmtheanswer$$$$ofQ95846.$$

Answered by mahdi last updated on 06/Jun/20

$$\Rightarrow 2^{\mathrm{x}}\times 1\mathrm{n}2+2\mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=2\mathrm{y}+2\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\Rightarrow$$$$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\times (2\mathrm{y}-2\mathrm{x})=2\mathrm{y}-2^{\mathrm{x}}.\ln 2\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2\mathrm{y}-2^{\mathrm{x}}.\ln 2}{2\mathrm{y}-2\mathrm{x}}$$

Answered by Sourav mridha last updated on 06/Jun/20

$$\Rightarrow 2^{\mathit{x}}\mathit{l}\mathit{n}\left(2\right)+2\mathit{y}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=2\mathit{y}+2\mathit{x}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$$$$\Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{2\mathit{y}-2^{\mathit{x}}\mathit{l}\mathit{n}\left(2\right)}{2\mathit{y}-2\mathit{x}}=\frac{2^{\mathit{x}}[1-\mathit{x}\mathit{l}\mathit{n}\left(2\right)]+{\mathbf{y}}^2}{2^{\mathbf{x}}+{\mathbf{y}}^2-2{\mathit{x}}^2}.$$

Answered by mathmax by abdo last updated on 06/Jun/20

$$\mathrm{e}\Rightarrow {\mathrm{y}}^2-2\mathrm{xy}+2^{\mathrm{x}}=0\to {\mathrm{\Delta }}^{{}^{\prime }}={\mathrm{x}}^2-2^{\mathrm{x}}\Rightarrow \mathrm{y}=\mathrm{x}\stackrel{-}{+}\sqrt{{\mathrm{x}}^2-2^{\mathrm{x}}}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=1\stackrel{-}{+}\frac{{({\mathrm{x}}^2-2^{\mathrm{x}})}^{{}^{\prime }}}{2\sqrt{{\mathrm{x}}^2-2^{\mathrm{x}}}}=1\stackrel{-}{+}\frac{2\mathrm{x}-\ln 2.2^{\mathrm{x}}}{2\sqrt{{\mathrm{x}}^2-2^{\mathrm{x}}}}$$

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