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

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

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$2$

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(B) Given $y = (	an^{-1} x)^2$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = 2(	an^{-1} x) \cdot \frac{1}{1+x^2}$. Multiplying both sides by $(1+x^2)$,we have: $(1+x^2) \frac{dy}{dx} = 2 	an^{-1} x$. Differentiating again with respect to $x$ using the product rule on the left side: $(1+x^2) \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot (2x) = 2 \cdot \frac{1}{1+x^2}$. Multiplying the entire equation by $(1+x^2)$,we get: $(1+x^2)^2 \frac{d^2 y}{dx^2} + 2x(1+x^2) \frac{dy}{dx} = 2$. Thus,the value is $2$.

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

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