Answer
$$ y'
=\frac{t}{\sqrt{t^2-9}}\sec\sqrt{t^2-9}\tan\sqrt{t^2-9}$$
Work Step by Step
Recall that $(\sec x)'=\sec x\tan x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\sec\sqrt{t^2-9}$, by using the chain rule, the derivative $ y'$ is given by
$$ y'=\sec\sqrt{t^2-9}\tan\sqrt{t^2-9} (\sqrt{t^2-9})'\\
=\frac{2t}{2\sqrt{t^2-9}}\sec\sqrt{t^2-9}\tan\sqrt{t^2-9}\\
=\frac{t}{\sqrt{t^2-9}}\sec\sqrt{t^2-9}\tan\sqrt{t^2-9}.$$
