Solveeit Logo
Login

Question

Question: If y = x<sup>2</sup> + \(\frac{1}{x^{2} + \frac{1}{x^{2} + ....\infty}}\), then \(\frac{dy}{dx}\) =...

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

A

2xy2y2+1\frac{2xy^{2}}{y^{2} + 1}

B

2xyy2+1\frac{2xy}{y^{2} + 1}

C

2xy2+1\frac{2x}{y^{2} + 1}

D

2y2+1\frac{2}{y^{2} + 1}

Answer

2xy2y2+1\frac{2xy^{2}}{y^{2} + 1}

Explanation

Solution

y = x2 + 1/y Ž y2 = x2y + 1 Ž y2 – x2y – 1 = 0

dydx=−(−2xy2y−x2)\frac{dy}{dx} = - \left( \frac{- 2xy}{2y - x^{2}} \right)= –(−2xy2y−(y−1/y))\left( \frac{- 2xy}{2y - (y - 1/y)} \right)= 2xy2y2+1\frac{2xy^{2}}{y^{2} + 1}