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

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Question

(A) Given the infinite series $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}$,we can observe that the expression repeats itself after the fir

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the infinite series $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}$,we can observe that the expression repeats itself after the first $x^2$. Thus,we can write the equation as $y = x^2 + \frac{1}{y}$. Multiplying both sides by $y$,we get $y^2 = x^2y + 1$. Differentiating both sides with respect to $x$ using the product rule: $2y \frac{dy}{dx} = x^2 \frac{dy}{dx} + y(2x)$. 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} = $

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