Find the second order derivative of the funct | vedclass

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Question

(A) Let $y = 	an^{-1} x$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = \frac{d}{dx}(	an^{-1} x) = \frac{1}{1+x^2}$. Now,to find the

Vedclass · Accepted Answer

$-\frac{2x}{(1+x^2)^2}$

Vedclass · Answer

(A) Let $y = 	an^{-1} x$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = \frac{d}{dx}(	an^{-1} x) = \frac{1}{1+x^2}$. Now,to find the second order derivative,we differentiate $\frac{dy}{dx}$ with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{1+x^2}ight) = \frac{d}{dx}(1+x^2)^{-1}$. Using the chain rule: $\frac{d^2y}{dx^2} = -1 \cdot (1+x^2)^{-2} \cdot \frac{d}{dx}(1+x^2)$. $\frac{d^2y}{dx^2} = -1 \cdot (1+x^2)^{-2} \cdot (2x)$. Therefore,$\frac{d^2y}{dx^2} = -\frac{2x}{(1+x^2)^2}$.

Find the second order derivative of the function $\tan^{-1} x$.

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