Question

If \( y\sqrt{1-x^2} + x\sqrt{1-y^2} = 1 \), then \( \dfrac{dy}{dx} = \)

• A. \( -\sqrt{\dfrac{1-y^2}{1-x^2}} \)
• B. \( -\sqrt{\dfrac{1-x^2}{1-y^2}} \)
• C. \( \sqrt{\dfrac{1+y^2}{1+x^2}} \)
• D. \( \sqrt{\dfrac{1-x^2}{1-y^2}} \)

Hint:

In implicit differentiation involving radicals, simplify using the original equation wherever possible.

Answer 1

The Correct Option is A

Step 1: Differentiate implicitly with respect to \( x \).
\[ \frac{d}{dx}\left(y\sqrt{1-x^2}\right) + \frac{d}{dx}\left(x\sqrt{1-y^2}\right) = 0 \] Step 2: Apply product rule.
\[ \sqrt{1-x^2}\frac{dy}{dx} - \frac{xy}{\sqrt{1-x^2}} + \sqrt{1-y^2} - \frac{x y}{\sqrt{1-y^2}}\frac{dy}{dx} = 0 \] Step 3: Collect \( \dfrac{dy}{dx} \) terms.
\[ \left(\sqrt{1-x^2} - \frac{x y}{\sqrt{1-y^2}}\right)\frac{dy}{dx} = \frac{x y}{\sqrt{1-x^2}} - \sqrt{1-y^2} \] Step 4: Simplify using the given equation.
\[ \frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}} \] Step 5: Conclusion.
Hence, \[ \frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}} \]