If 1-x^6 + 1-y^6 =a^3(x^3-y)^3, then dy dx is equal to:

MCQ

If $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}^3(\text{x}^3-\text{y})^3,$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:

✓
$\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$
B
$\frac{\text{y}^2}{\text{x}^2}\sqrt{\frac{1-\text{y}^6}{1+\text{x}^6}}$
C
$\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{x}^6}{1-\text{y}^6}}$
D
$\text{None of these.}$

✓

Answer

Correct option: A.

$\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$

We have, $\sqrt{1-\text{x}^6}+\sqrt{1-\text{y}^6}=\text{a}(\text{x}^3-\text{y}^3)$
Putting $\text{x}^3=\sin\text{A}\text{ and y}^3=\sin\text{B}$
$\Rightarrow\sqrt{1-\sin^2\text{A}}+\sqrt{1-\sin2\text{B}}=\text{a}(\sin\text{A}-\sin\text{B})$
$\Rightarrow\cos\text{A}+\cos\text{B}=\text{a}(\sin\text{A}-\sin\text{B})$
$\Rightarrow2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{b}}{2}\Big) \\ =2\text{a}\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)$
$\Rightarrow\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\text{a}$
$\Rightarrow\frac{\text{A}+\text{B}}{2}=\cot^{-1}(\text{a})$
$\Rightarrow\text{A}+\text{B}=2\cot^{-1}(\text{a})$
$\Rightarrow\sin^{-1}\text{x}^{3}-\sin^{-1}\text{y}^3=2\cot^{-1}(\text{a})$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}\times\frac{\text{d}}{\text{dx}}(\text{x}^3)-\frac{1}{\sqrt{1-\text{y}^6}}\times\frac{\text{d}}{\text{dx}}(\text{y}^3)=0$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^6}}3\text{x}^2-\frac{1}{\sqrt{1-\text{y}^6}}\times3\text{y}^2\times\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{\text{y}^2}\sqrt{\frac{1-\text{y}^6}{1-\text{x}^6}}$