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Question: If \(y = x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + ......\infty}}}\), then \(\frac{...

If y=x2+1x2+1x2+1x2+......∞y = x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + \frac{1}{x^{2} + ......\infty}}}, then dydx=\frac{dy}{dx} =

A

2xy2y−x2\frac{2xy}{2y - x^{2}}

B

xyy+x2\frac{xy}{y + x^{2}}

C

xyy−x2\frac{xy}{y - x^{2}}

D

2x2+x2y\frac{2x}{2 + \frac{x^{2}}{y}}

Answer

2xy2y−x2\frac{2xy}{2y - x^{2}}

Explanation

Solution

y=x2+1yy = x^{2} + \frac{1}{y} ⇒ y2=x2y+1y^{2} = x^{2}y + 1 ⇒ 2ydydx=y.2x+x2dydx2y\frac{dy}{dx} = y.2x + x^{2}\frac{dy}{dx}

⇒ dydx=2xy2y−x2\frac{dy}{dx} = \frac{2xy}{2y - x^{2}}