If y = e^{tan^{-1}x},then (1 + x^2)frac{d^2y} | vedclass

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(A) Given $y = e^{	an^{-1}x}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = e^{	an^{-1}x} \cdot \frac{1}{1+x^2} = \frac{y}{1+x^2}$.This i

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$(1 - 2x)\frac{dy}{dx}$

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(A) Given $y = e^{	an^{-1}x}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = e^{	an^{-1}x} \cdot \frac{1}{1+x^2} = \frac{y}{1+x^2}$.This implies $(1+x^2)\frac{dy}{dx} = y$.Differentiating both sides with respect to $x$ using the product rule:$(1+x^2)\frac{d^2y}{dx^2} + \frac{dy}{dx}(2x) = \frac{dy}{dx}$.Rearranging the terms:$(1+x^2)\frac{d^2y}{dx^2} = \frac{dy}{dx} - 2x\frac{dy}{dx}$.$(1+x^2)\frac{d^2y}{dx^2} = (1-2x)\frac{dy}{dx}$.

If $y = e^{\tan^{-1}x}$,then $(1 + x^2)\frac{d^2y}{dx^2} = $

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