Correct Answer - Option 3 :
\(\frac{\sqrt{1 - y^{2}}}{1 - 2y^{2}}\)
Concept:
\(\frac{\mathrm{d} (uv)}{\mathrm{d} x}\) = \(u\frac{\mathrm{d} v}{\mathrm{d} x} + v\frac{\mathrm{d}u }{\mathrm{d} x}\)
Calculation:
Given that x = y \(\sqrt{1 - \rm y^{2}}\)
Differentiating with respect to x, we get
\(\frac{\mathrm{d} \rm x}{\mathrm{d} x}\) = \(\frac{\mathrm{d} \left (\rm y\sqrt{1 -\rm y^{2}} \right )}{\mathrm{d} x}\)
⇒ 1 = \(\rm \rm \sqrt{1 -\rm y^{2}} \frac{\mathrm{d}y}{\mathrm{d} x}\) + \(\frac{\rm y\mathrm{d} \sqrt{1 - y^{2}}}{\mathrm{d} x}\)
⇒ 1 = \(\sqrt{1-\rm y^{2}}\) \(\frac{\mathrm{d} \rm y}{\mathrm{d} x}\) + y\(\frac{1}{2\sqrt{1 -\rm y^{2}}}\)( - 2y)\(\frac{\mathrm{d} \rm y}{\mathrm{d} x}\)
⇒ 1 = \(\frac{\mathrm{d} \rm y}{\mathrm{d} x}\)\(\left [ \frac{1 - y^{2} - y^{2}}{\sqrt{1 - y^{2}}} \right ]\)
⇒ 1 = \(\left [ \frac{1 - 2y^{2} }{\sqrt{1 - y^{2}}} \right ]\)\(\frac{\mathrm{d} y}{\mathrm{d} x}\)
∴ \(\frac{\mathrm{d} y}{\mathrm{d} x}\) = \(\left [ \frac{\sqrt{1 - y^{2}}}{1 - 2y^{2}} \right ]\)