Question

If $\sqrt{1-x^{6} }+ \sqrt{1-y^{6}} =a\left(x^{3} -y^{3}\right) , $ then $ y^{2} \frac{dy}{dx} = $

• A. $\sqrt{\frac{ 1- y^6}{ 1 - x^6}}$
• B. $ x \sqrt{\frac{ 1- y^6}{ 1 - x^6}}$
• C. $x^2 \sqrt{\frac{ 1- y^6}{ 1 - x^6}}$
• D. $\frac{1}{x^2} \sqrt{\frac{ 1- y^6}{ 1 - x^6}}$

Answer 1

Answer: (C)

Given equation is
$\sqrt{1-x^{6}}+\sqrt{1-y^{6}}=a\left(x^{3}-y^{3}\right)$
$\Rightarrow \frac{\sqrt{1-x^{6}}+\sqrt{1-y^{6}}}{x^{3}-y^{3}}=a$
On differentiating both sides w.r.t., $x$, we get
$\frac{\left[\left(x^{3}-y^{3}\right) \left(\frac{-6x^{5}}{2\sqrt{1 -x^{6}}} -\frac{6y^{5}}{2\sqrt{1 -y^{6}}} \frac{dy}{dx}\right) - \left(\sqrt{1 -x^{6}}+\sqrt{1- y^{6}}\right)\left(3x^{2} -3y^{2} \frac{dy}{dx}\right)\right]}{\left(x^{3}-y^{3}\right)^{2}}=0$
$\Rightarrow \left(y^{2}\left(\sqrt{1-x^{6}}+\sqrt{1-y^{6}}\right)-\frac{y^{5}\left(x^{3}-y^{3}\right)}{\sqrt{1-y^{6}}}\right) \frac{d y}{d x}$
$=x^{2}\left(\sqrt{1-x^{6}}+\sqrt{1-y^{6}}\right)+\frac{x^{5}}{\sqrt{1-x^{6}}}\left(x^{3}-y^{3}\right)$
$\Rightarrow y^{2} \frac{d y}{d x}\left[\frac{\sqrt{1-x^{6}}+\sqrt{1-y^{6}}+\left(1-y^{6}\right)-y^{3} x^{3}+y^{6}}{\sqrt{1-y^{6}}}\right]$
$=x^{2}\left[\frac{\left(1-x^{6}\right)+\sqrt{1-y^{6}} \sqrt{1-x^{6}}+x^{6}-x^{3} y^{3}}{\sqrt{1-x^{6}}}\right]$
$\Rightarrow y^{2} \frac{d y}{d x}\left[\frac{\sqrt{1-x^{6}} \sqrt{1-y^{6}}+1-x^{3} y^{3}}{\sqrt{1-y^{6}}}\right]$
$=x^{2}\left[\frac{1+\sqrt{1-y^{6}} \sqrt{1-x^{6}}-x^{3} y^{3}}{\sqrt{1-x^{6}}}\right]$
$\Rightarrow y^{2} \frac{d y}{d x}=x^{2} \sqrt{\frac{1-y^{6}}{1-x^{6}}}$