If y = (1 - x^2) operatorname{Tanh}^{-1} x, t | vedclass

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(D) Given $y = (1 - x^2) \operatorname{Tanh}^{-1} x$. First, differentiate with respect to $x$: $\frac{dy}{dx} = (1 - x^2) \cdot \frac{d}{dx}(\operato

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $y = (1 - x^2) \operatorname{Tanh}^{-1} x$. First, differentiate with respect to $x$: $\frac{dy}{dx} = (1 - x^2) \cdot \frac{d}{dx}(\operatorname{Tanh}^{-1} x) + \operatorname{Tanh}^{-1} x \cdot \frac{d}{dx}(1 - x^2)$ $\frac{dy}{dx} = (1 - x^2) \cdot \frac{1}{1 - x^2} + \operatorname{Tanh}^{-1} x \cdot (-2x)$ $\frac{dy}{dx} = 1 - 2x \operatorname{Tanh}^{-1} x$. Now, differentiate again with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{d}{dx}(1) - 2 \cdot \frac{d}{dx}(x \operatorname{Tanh}^{-1} x)$ $\frac{d^2 y}{dx^2} = 0 - 2 \left[ x \cdot \frac{1}{1 - x^2} + \operatorname{Tanh}^{-1} x \cdot 1 \right]$ $\frac{d^2 y}{dx^2} = -2 \left[ \frac{x}{1 - x^2} + \operatorname{Tanh}^{-1} x \right]$. From the original equation, $\operatorname{Tanh}^{-1} x = \frac{y}{1 - x^2}$. Substituting this into the expression: $\frac{d^2 y}{dx^2} = -2 \left[ \frac{x}{1 - x^2} + \frac{y}{1 - x^2} \right]$ $\frac{d^2 y}{dx^2} = -\frac{2(x + y)}{1 - x^2}$.

If $y = (1 - x^2) \operatorname{Tanh}^{-1} x$, then $\frac{d^2 y}{d x^2} = $

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