If y = x^2 + frac{1}{x^2 + frac{1}{x^2 + frac | vedclass

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(A) Given the infinite series $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}$.Since the series is infinite,we can observe that the expression

Vedclass · Accepted Answer

$\frac{2xy}{2y - x^2}$

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(A) Given the infinite series $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}$.Since the series is infinite,we can observe that the expression repeats itself,so we can write: $y = x^2 + \frac{1}{y}$.Multiplying both sides by $y$,we get: $y^2 = x^2 y + 1$.Differentiating both sides with respect to $x$ using the product rule and chain rule:$2y \frac{dy}{dx} = (x^2 \frac{dy}{dx} + y \cdot 2x) + 0$.Rearranging the terms to isolate $\frac{dy}{dx}$:$2y \frac{dy}{dx} - x^2 \frac{dy}{dx} = 2xy$.$\frac{dy}{dx} (2y - x^2) = 2xy$.Therefore,$\frac{dy}{dx} = \frac{2xy}{2y - x^2}$.

If $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}}$,then $\frac{dy}{dx}$ is equal to

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