If x = e^{tan^{-1}left(frac{y-x^2}{x^2}right) | vedclass

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(D) Given the equation $x = e^{	an^{-1}\left(\frac{y-x^2}{x^2}ight)}$.Taking the natural logarithm on both sides,we get $\ln(x) = 	an^{-1}\left(\f

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$3$

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(D) Given the equation $x = e^{	an^{-1}\left(\frac{y-x^2}{x^2}ight)}$.Taking the natural logarithm on both sides,we get $\ln(x) = 	an^{-1}\left(\frac{y-x^2}{x^2}ight)$.This implies $	an(\ln x) = \frac{y-x^2}{x^2}$.Rearranging the terms,we have $y = x^2 	an(\ln x) + x^2$.Differentiating both sides with respect to $x$ using the product rule and chain rule:$\frac{dy}{dx} = \frac{d}{dx}[x^2 	an(\ln x)] + \frac{d}{dx}[x^2]$$\frac{dy}{dx} = [2x 	an(\ln x) + x^2 \cdot \sec^2(\ln x) \cdot \frac{1}{x}] + 2x$$\frac{dy}{dx} = 2x 	an(\ln x) + x \sec^2(\ln x) + 2x$.Now,evaluate at $x = 1$:$\frac{dy}{dx} \Big|_{x=1} = 2(1) 	an(\ln 1) + 1 \sec^2(\ln 1) + 2(1)$Since $\ln 1 = 0$ and $	an 0 = 0$ and $\sec 0 = 1$:$\frac{dy}{dx} \Big|_{x=1} = 2(0) + 1(1)^2 + 2 = 0 + 1 + 2 = 3$.

If $x = e^{\tan^{-1}\left(\frac{y-x^2}{x^2}\right)}$,then $\frac{dy}{dx}$ at $x = 1$ is

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