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 If x = y \(\sqrt{1 - \rm y^{2}}\) , then \(\frac{\mathrm{d} \rm y}{\mathrm{d} x}\) is equal to

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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 = \(\left [ \frac{1 - 2y^{2} }{\sqrt{1 - y^{2}}} \right ]\)\(\frac{\mathrm{d} y}{\mathrm{d} x}\)

∴  \(\frac{\mathrm{d} y}{\mathrm{d} x}\) = \(\left [ \frac{\sqrt{1 - y^{2}}}{1 - 2y^{2}} \right ]\)  

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