Question

If $ y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+...\infty }}}} $ ,then $ \frac{dy}{dx} $ is equal to

• A. $ \frac{y+x}{{{y}^{2}}-2x} $
• B. $ \frac{{{y}^{3}}-x}{2{{y}^{2}}-2xy-1} $
• C. $ \frac{{{y}^{3}}+x}{2{{y}^{2}}-x} $
• D. None of these

Answer 1

Answer: (D)

We have, $ y=\sqrt{x+\sqrt{y+\sqrt{x+\sqrt{y+...\infty }}}} $ $ \Rightarrow $ $ {{y}^{2}}=x+\sqrt{y+\sqrt{x+\sqrt{y+...+\infty }}} $ $ \Rightarrow $ $ {{y}^{2}}=x+\sqrt{y+y} $ $ \Rightarrow $ $ ({{y}^{2}}-{{x}^{2}})=2y $ On differentiating both sides w. r. t . $ x $ , we get $ 2({{y}^{2-x}})\left( 2y\frac{dy}{dx}-1 \right)=2\frac{dy}{dx} $ $ \Rightarrow $ $ 2y({{y}^{2}}-x)\frac{dy}{dx}-\frac{dy}{dx}={{y}^{2}}-x $ $ \Rightarrow $ $ \frac{dy}{dx}(2{{y}^{3}}-2xy-1)=({{y}^{2}}-x) $ $ \Rightarrow $ $ \frac{dy}{dx}=\frac{{{y}^{2}}-x}{(2{{y}^{3}}-2xy-1)} $