Find (d y)/(d x) if operatorname(Sec)^-1(x^2-1)

To find \( \frac{dy}{dx} \) for \( \operatorname{Sec}^{-1}\left(x^{2}-1\right) \), we can use the chain rule and the derivative of the inverse secant function.<br /><br />The derivative of the inverse secant function is given by:<br /><br />\[ \frac{d}{dx}(\operatorname{Sec}^{-1}(u)) = -\frac{1}{|u|\sqrt{u^2-1}} \]<br /><br />Now, let's apply the chain rule to find \( \frac{dy}{dx} \):<br /><br />\[ \frac{dy}{dx} = \frac{d}{dx}(\operatorname{Sec}^{-1}(x^{2}-1)) \]<br /><br />Using the chain rule, we have:<br /><br />\[ \frac{dy}{dx} = \frac{d}{du}(\operatorname{Sec}^{-1}(u)) \cdot \frac{du}{dx} \]<br /><br />where \( u = x^{2}-1 \).<br /><br />Now, let's find \( \frac{du}{dx} \):<br /><br />\[ \frac{du}{dx} = \frac{d}{dx}(x^{2}-1) = 2x \]<br /><br />Now, substitute \( u = x^{2}-1 \) into the derivative of the inverse secant function:<br /><br />\[ \frac{d}{dx}(\operatorname{Sec}^{-1}(x^{2}-1)) = -\frac{1}{|x^{2}-1|\sqrt{(x^{2}-1)^{2}-1}} \cdot 2x \]<br /><br />Simplifying further, we get:<br /><br />\[ \frac{dy}{dx} = -\frac{2x}{|x^{2}-1|\sqrt{x^{4}-2x^{2}+2}} \]<br /><br />Therefore, the derivative of \( \operatorname{Sec}^{-1}\left(x^{2}-1\right) \) with respect to \( x \) is:<br /><br />\[ \frac{dy}{dx} = -\frac{2x}{|x^{2}-1|\sqrt{x^{4}-2x^{2}+2}} \]