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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

(√(1−x^6 ))+(√(1−y^6 ))=9(x^3 −y^3 )  ⇒((−6x^5 )/(2(√(1−x^6 ))))+((−6y^5 )/(2(√(1−y^6 ))))×(dy/dx)=27x^2 −27y^2 (dy/dx)  ⇒(((−3y^5 )/(√(1−y^6 )))+27y^2 )(dy/dx)=(27x^2 +((3x^5 )/(√(1−x^6 ))))  ⇒(dy/dx)=(x^2 /y^2 )((((9(√(1−x^6 ))+x^3 )(√(1−y^6 )))/((9(√(1−y^6 ))−y^3 )(√(1−x^6 )))))  ≠(x^2 /y^2 )(((√(1−y^2 ))/(√(1−x^2 ))))  please recheck the question

$$\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }+\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }=\mathrm{9}\left({x}^{\mathrm{3}} −{y}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\frac{−\mathrm{6}{x}^{\mathrm{5}} }{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }}+\frac{−\mathrm{6}{y}^{\mathrm{5}} }{\mathrm{2}\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }}×\frac{{dy}}{{dx}}=\mathrm{27}{x}^{\mathrm{2}} −\mathrm{27}{y}^{\mathrm{2}} \frac{{dy}}{{dx}} \\ $$$$\Rightarrow\left(\frac{−\mathrm{3}{y}^{\mathrm{5}} }{\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }}+\mathrm{27}{y}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}=\left(\mathrm{27}{x}^{\mathrm{2}} +\frac{\mathrm{3}{x}^{\mathrm{5}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }}\right) \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\left(\frac{\left(\mathrm{9}\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }+{x}^{\mathrm{3}} \right)\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }}{\left(\mathrm{9}\sqrt{\mathrm{1}−{y}^{\mathrm{6}} }−{y}^{\mathrm{3}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{6}} }}\right) \\ $$$$\neq\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\left(\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right) \\ $$$${please}\:{recheck}\:{the}\:{question} \\ $$

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