Answer:
[tex]y' = \( \frac{12s^2}{|4s^3 + 9|\sqrt{16s^6 + 72s^3 + 80}} \)[/tex]
Step-by-step explanation:
We will need to use the inverse trig function derivative identity for secant, and plug it into the formula to find the derivative.
[tex]~~~~~~~~~~~~~~~~~\textbf{Inverse Trig Derivative Identities}[/tex]
[tex]\fbox{\begin{minipage}{\textwidth}\begin{align*}\frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} \\\\\frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1-x^2}} \\\\\frac{d}{dx}(\arctan(x)) = \frac{1}{1+x^2} \\\\\boxed{\frac{d}{dx}(\text{arcsec}(x)) = \frac{1}{|x|\sqrt{x^2-1}} }\\\\\frac{d}{dx}(\text{arccsc}(x)) = -\frac{1}{|x|\sqrt{x^2-1}} \\\\\frac{d}{dx}(\text{arccot}(x)) = -\frac{1}{1+x^2}\end{align*}\end{minipage}}[/tex]
We must use the arcsec(x) derivative identity to solve this.
Solving:
[tex]\( \frac{d}{dx}(\sec^{-1}(u)) = \frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx} \)[/tex]
[tex]\( u = 4s^3 + 9 \)[/tex]
[tex]\( \frac{du}{ds} = 12s^2 \)[/tex]
[tex]\noindent\makebox[\linewidth]{\rule{\textwidth}{0.4pt}}[/tex]
[tex]\frac{d}{ds}\left(\sec^{-1}(4s^3 + 9)\right) = \frac{1}{|4s^3 + 9|\sqrt{(4s^3 + 9)^2 - 1}} \cdot \frac{du}{ds}[/tex]
[tex]y' = \frac{1}{|4s^3 + 9|\sqrt{(16s^6 + 72s^3 + 81) - 1}} \cdot 12s^2[/tex]
[tex]\therefore{\boxed{y'= \frac{12s^2}{|4s^3 + 9|\sqrt{16s^6 + 72s^3 + 80}}}[/tex]
