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Question Number 62102 by rajesh4661kumar@gamil.com last updated on 15/Jun/19

Answered by Kunal12588 last updated on 15/Jun/19

$$\sqrt{1-x^6}+\sqrt{1-y^6}=9(x^3-y^3)$$$$\Rightarrow \frac{-6x^5}{2\sqrt{1-x^6}}+\frac{-6y^5}{2\sqrt{1-y^6}}\times \frac{dy}{dx}=27x^2-27y^2\frac{dy}{dx}$$$$\Rightarrow (\frac{-3y^5}{\sqrt{1-y^6}}+27y^2)\frac{dy}{dx}=(27x^2+\frac{3x^5}{\sqrt{1-x^6}})$$$$\Rightarrow \frac{dy}{dx}=\frac{x^2}{y^2}\left(\frac{(9\sqrt{1-x^6}+x^3)\sqrt{1-y^6}}{(9\sqrt{1-y^6}-y^3)\sqrt{1-x^6}}\right)$$$$\ne \frac{x^2}{y^2}\left(\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\right)$$$$pleaserecheckthequestion$$

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