If y = e^{x^2 + e^{x^2 + e^{x^2} + dots}} the | vedclass

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(C) Given the equation $y = e^{x^2 + e^{x^2 + e^{x^2} + \dots}}$.Since the exponent repeats infinitely,we can write $y = e^{x^2 + y}$.Taking the natur

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$\frac{2xy}{1 - y}$

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(C) Given the equation $y = e^{x^2 + e^{x^2 + e^{x^2} + \dots}}$.Since the exponent repeats infinitely,we can write $y = e^{x^2 + y}$.Taking the natural logarithm on both sides,we get $\ln(y) = x^2 + y$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(\ln(y)) = \frac{d}{dx}(x^2 + y)$$\frac{1}{y} \frac{dy}{dx} = 2x + \frac{dy}{dx}$Rearranging the terms to solve for $\frac{dy}{dx}$:$\frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} = 2x$$\frac{dy}{dx} (\frac{1}{y} - 1) = 2x$$\frac{dy}{dx} (\frac{1 - y}{y}) = 2x$$\frac{dy}{dx} = \frac{2xy}{1 - y}$.

If $y = e^{x^2 + e^{x^2 + e^{x^2} + \dots}}$ then $\frac{dy}{dx} = $

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