If x+y=tan ^{-1} y and frac{d^{2} y}{d x^{2}} | vedclass

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(B) Given,$x+y=	an ^{-1} y$.Differentiating both sides with respect to $x$,we get:$1 + \frac{dy}{dx} = \frac{1}{1+y^2} \cdot \frac{dy}{dx}$Rearrangin

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$\frac{2}{y^{3}}$

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(B) Given,$x+y=	an ^{-1} y$.Differentiating both sides with respect to $x$,we get:$1 + \frac{dy}{dx} = \frac{1}{1+y^2} \cdot \frac{dy}{dx}$Rearranging the terms:$\frac{dy}{dx} \left( \frac{1}{1+y^2} - 1 ight) = 1$$\frac{dy}{dx} \left( \frac{1 - 1 - y^2}{1+y^2} ight) = 1$$\frac{dy}{dx} \left( \frac{-y^2}{1+y^2} ight) = 1$$\frac{dy}{dx} = -\frac{1+y^2}{y^2} = -\left( \frac{1}{y^2} + 1 ight)$Now,differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = -\frac{d}{dx} (y^{-2} + 1)$$\frac{d^2y}{dx^2} = -(-2y^{-3}) \cdot \frac{dy}{dx}$$\frac{d^2y}{dx^2} = \frac{2}{y^3} \cdot \frac{dy}{dx}$Comparing this with the given equation $\frac{d^2y}{dx^2} = f(y) \frac{dy}{dx}$,we get $f(y) = \frac{2}{y^3}$.

If $x+y=\tan ^{-1} y$ and $\frac{d^{2} y}{d x^{2}}=f(y) \frac{d y}{d x}$,then $f(y)$ is equal to

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