find the derivative of y with respect to s.\n\n$y = \\text{arcsec}(2s^{4}+3)$\n\n$\\frac{dy}{ds}=\\square$

Explanation:

Step1: Recall the derivative formula for (y = \text{arcsec}(u))

The derivative of (y=\text{arcsec}(u)) with respect to (s) is (\frac{dy}{ds}=\frac{u'}{|u|\sqrt{u^{2}-1}}), where (u = 2s^{4}+3) and (u') is the derivative of (u) with respect to (s).

Step2: Find the derivative of (u = 2s^{4}+3)

Using the power rule ((x^{n})'=nx^{n - 1}), we have (u'=\frac{d}{ds}(2s^{4}+3)=8s^{3}).

Step3: Substitute (u) and (u') into the formula

Since (u = 2s^{4}+3), and for (s) such that (|2s^{4}+3|=2s^{4}+3) (because (2s^{4}+3>0) for all real (s) as (s^{4}\geq0)), we substitute into (\frac{dy}{ds}=\frac{u'}{|u|\sqrt{u^{2}-1}}). [ \begin{align*} \frac{dy}{ds}&=\frac{8s^{3}}{(2s^{4}+3)\sqrt{(2s^{4}+3)^{2}-1}}\ &=\frac{8s^{3}}{(2s^{4}+3)\sqrt{4s^{8}+12s^{4}+9 - 1}}\ &=\frac{8s^{3}}{(2s^{4}+3)\sqrt{4s^{8}+12s^{4}+8}}\ &=\frac{8s^{3}}{(2s^{4}+3)\sqrt{4(s^{8}+3s^{4}+2)}}\ &=\frac{8s^{3}}{(2s^{4}+3)\times2\sqrt{s^{8}+3s^{4}+2}}\ &=\frac{4s^{3}}{(2s^{4}+3)\sqrt{(s^{4}+1)(s^{4}+2)}} \end{align*} ]

Answer:

(\frac{4s^{3}}{(2s^{4}+3)\sqrt{(s^{4}+1)(s^{4}+2)}})