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Question Number 148324 by Odhiambojr last updated on 27/Jul/21

$$x^2-y^2=3,finddy/dx$$

Answered by liberty last updated on 27/Jul/21

$$2\mathrm{x}-{2\mathrm{yy}}^{\prime }=0$$$$\Rightarrow {\mathrm{y}}^{\prime }=\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{x}}{\pm \sqrt{{\mathrm{x}}^2-3}}$$

Answered by mathmax by abdo last updated on 27/Jul/21

$${\mathrm{x}}^2-{\mathrm{y}}^2=3\Rightarrow {\mathrm{x}}^2+3={\mathrm{y}}^2\Rightarrow \mathrm{y}=\stackrel{-}{+}\sqrt{{\mathrm{x}}^2+3}\Rightarrow$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\stackrel{-}{+}\frac{2\mathrm{x}}{2\sqrt{{\mathrm{x}}^2+3}}=\stackrel{-}{+}\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+3}}$$

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