If y = operatorname{Sec}^{-1} x, then frac{d^ | vedclass

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(A) Given $y = \operatorname{Sec}^{-1} x$. First, differentiate with respect to $x$: $\frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}$. This can be written

Vedclass · Accepted Answer

$\frac{1-2 x^2}{x|x|\left(x^2-1\right)^{\frac{3}{2}}}$

Vedclass · Answer

(A) Given $y = \operatorname{Sec}^{-1} x$. First, differentiate with respect to $x$: $\frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}$. This can be written as $\frac{dy}{dx} = (x^2-1)^{-\frac{1}{2}} \cdot \frac{1}{|x|}$. For $x > 1$, $|x| = x$, so $\frac{dy}{dx} = x^{-1}(x^2-1)^{-\frac{1}{2}}$. Differentiating again using the product rule: $\frac{d^2y}{dx^2} = \frac{d}{dx} [x^{-1}(x^2-1)^{-\frac{1}{2}}] = -x^{-2}(x^2-1)^{-\frac{1}{2}} + x^{-1} \cdot (-\frac{1}{2})(x^2-1)^{-\frac{3}{2}} \cdot (2x)$. $\frac{d^2y}{dx^2} = -\frac{1}{x^2\sqrt{x^2-1}} - \frac{1}{(x^2-1)^{\frac{3}{2}}} = \frac{-(x^2-1) - x^2}{x^2(x^2-1)^{\frac{3}{2}}} = \frac{1-2x^2}{x^2(x^2-1)^{\frac{3}{2}}}$. Considering the absolute value for all $x$, the general form is $\frac{1-2x^2}{x|x|(x^2-1)^{\frac{3}{2}}}$. Thus, the correct option is $A$.

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

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