Find (d y)/(d x) if [sec ^-1(x^2-1)]^5 ( mart )

To find \( \frac{dy}{dx} \) for the given function \( \left[\sec^{-1}\left(x^2 - 1\right)\right]^5 \), we can use the chain rule and the derivative of the inverse secant function.<br /><br />Let's define \( u = \sec^{-1}(x^2 - 1) \). Then, \( y = u^5 \).<br /><br />First, we need to find \( \frac{du}{dx} \). To do this, we'll differentiate the inner function \( \sec^{-1}(x^2 - 1) \) with respect to \( x \).<br /><br />The derivative of \( \sec^{-1}(u) \) with respect to \( u \) is \( \frac{1}{|u|\sqrt{u^2 - 1}} \). Now, let's find \( \frac{du}{dx} \) by applying the chain rule:<br /><br />\[ \frac{du}{dx} = \frac{d}{dx} \sec^{-1}(x^2 - 1) = \frac{1}{|(x^2 - 1)|\sqrt{(x^2 - 1)^2 - 1}} \cdot \frac{d}{dx}(x^2 - 1) \]<br /><br />\[ \frac{du}{dx} = \frac{1}{|(x^2 - 1)|\sqrt{x^4 - 2x^2 + 1 - 1}} \cdot 2x \]<br /><br />\[ \frac{du}{dx} = \frac{2x}{|(x^2 - 1)|\sqrt{x^4 - 2x^2}} \]<br /><br />Now, let's find \( \frac{dy}{du} \) by differentiating \( y = u^5 \) with respect to \( u \):<br /><br />\[ \frac{dy}{du} = 5u^4 \]<br /><br />Finally, we can find \( \frac{dy}{dx} \) by applying the chain rule:<br /><br />\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 5u^4 \cdot \frac{2x}{|(x^2 - 1)|\sqrt{x^4 - 2x^2}} \]<br /><br />Substituting back \( u = \sec^{-1}(x^2 - 1) \), we get:<br /><br />\[ \frac{dy}{dx} = 5 \left(\sec^{-1}(x^2 - 1)\right)^4 \cdot \frac{2x}{|(x^2 - 1)|\sqrt{x^4 - 2x^2}} \]<br /><br />Therefore, the derivative of the given function is:<br /><br />\[ \frac{dy}{dx} = 10x \cdot \frac{\left(\sec^{-1}(x^2 - 1)\right)^4}{|(x^2 - 1)|\sqrt{x^4 - 2x^2}} \]