Question

If $y\sqrt{1-x^{2}} + x\sqrt{1-y^{2}} = 1$ show that $\frac{dy}{dx} = -\sqrt{\frac{1-y^{2}}{1-x^{2}}}$

Answer 1

$\blue{\mathfrak{Here is your answer.......Hope this helps you....}}$

Answer 2

Answer:

We have,

$y\sqrt{1-x^{2}} + x\sqrt{1-y^{2}} = 1.$

To show:

$\dfrac{\text dy}{\text dx}=-\sqrt{\dfrac{1-y^2}{1-x^2}}.$

On differentiating both the sides with respect to x, we get,

$\dfrac{\text d}{\text dx}\left (y\sqrt{1-x^{2}} + x\sqrt{1-y^{2}} \right ) = \dfrac{\text d}{\text dx}(1)\\\sqrt {1-x^2}\dfrac{\text dy}{\text dx}+y\dfrac{\text d}{\text dx}(\sqrt{1-x^2})+\sqrt {1-y^2}\dfrac{\text dx}{\text dx}+x\dfrac{\text d}{\text dx}(\sqrt{1-y^2})=0\\\sqrt {1-x^2}\dfrac{\text dy}{\text dx}+\dfrac{y}{2\sqrt{1-x^2}}\dfrac{\text d}{\text dx}(1-x^2)+\sqrt {1-y^2}+\dfrac{x}{2\sqrt{1-y^2}}\dfrac{\text d}{\text dx}(1-y^2)=0$

$\sqrt {1-x^2}\dfrac{\text dy}{\text dx}+\dfrac{y}{2\sqrt{1-x^2}}(-2x)+\sqrt {1-y^2}+\dfrac{x}{2\sqrt{1-y^2}}(-2y)\dfrac{\text dy}{\text dx}=0\\\left (\sqrt {1-x^2}+\dfrac{x}{2\sqrt{1-y^2}}(-2y) \right )\dfrac{\text dy}{\text dx}+\dfrac{y}{2\sqrt{1-x^2}}(-2x)+\sqrt{1-y^2}=0\\\left (\sqrt {1-x^2}-\dfrac{xy}{\sqrt{1-y^2}} \right )\dfrac{\text dy}{\text dx}+\left( \sqrt{1-y^2}-\dfrac{xy}{\sqrt{1-x^2}} \right )=0$

$\left (\dfrac{\sqrt {1-x^2}\sqrt{1-y^2}-xy}{\sqrt{1-y^2}} \right )\dfrac{\text dy}{\text dx}+\left (\dfrac{\sqrt {1-y^2}\sqrt{1-x^2}-xy}{\sqrt{1-x^2}} \right )=0\\\Rightarrow \dfrac{\text dy}{\text dx}=\dfrac{-\left (\dfrac{\sqrt {1-y^2}\sqrt{1-x^2}-xy}{\sqrt{1-x^2}} \right )}{\left (\dfrac{\sqrt {1-y^2}\sqrt{1-x^2}-xy}{\sqrt{1-y^2}} \right )}=-\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=-\sqrt{\dfrac{1-y^2}{1-x^2}}.$

It is the final expression, which was to be shown.