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Question Number 205827 by mustafazaheen last updated on 31/Mar/24

$${\mathrm{x}}^3+{\mathrm{y}}^3=1$$$$\mathrm{find}\mathrm{the}\mathrm{implceat}\mathrm{second}\mathrm{derivative}$$

Answered by cortano12 last updated on 31/Mar/24

$$3x^2+3y^2{y}^{\prime }=0$$$$\frac{dy}{dx}=-\frac{x^2}{y^2}=-x^2{y}^{-2}$$$$\frac{d}{dx}\left[\frac{dy}{dx}\right]=\frac{d}{dx}[-x^2{y}^{-2}]$$$$\frac{d^{2}y}{{dx}^2}=-2{xy}^{-2}+2x^2{y}^{-3}{y}^{\prime }$$$$\frac{d^{2}y}{{dx}^2}=-\frac{2x}{y^2}+\frac{2x^2}{y^3}.\left(\frac{-x^2}{y^2}\right)$$$$=\frac{-2x}{y^2}-\frac{2x^4}{y^5}=\frac{-2{xy}^3-2x^4}{y^5}$$$$=\frac{-2x(x^3+y^3)}{y^5}=-\frac{2x}{y^5}$$

Answered by TonyCWX08 last updated on 31/Mar/24

$$$$$$x^3+y^3-1=0$$$$f(x,y)=x^3+y^3-1$$$$f_x(x,y)=3x^2$$$$f_y(x,y)=3y^2$$$$$$$$\frac{dy}{dx}$$$$=-\frac{3x^2}{3y^2}$$$$=-\frac{x^2}{y^2}$$$$$$$$f(x,y)=-\frac{x^2}{y^2}$$$$f_x(x,y)=-\frac{2x}{y^2}$$$$f_y(x,y)=\frac{2x^2}{y^3}$$$$\frac{d^{2}y}{{dx}^2}$$$$=-\frac{-2x}{y^2}\mathbf{÷}\frac{2x^2}{y^3}$$$$=\frac{2x}{y^2}\times \frac{y^3}{2x^2}$$$$=\frac{y}{x}$$

Commented by Frix last updated on 31/Mar/24

$$\mathrm{Obviously}\mathrm{this}\mathrm{is}\mathrm{wrong}.$$$$x^3+y^3=1\iff y={(1-x^3)}^{\frac{1}{3}}\Rightarrow$$$$y^{″}=-\frac{2x}{{(1-x^3)}^{\frac{5}{3}}}=-\frac{2x}{y^5}\ne \frac{y}{x}$$

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