Answer
$\frac{1}{4\sqrt{ 15}} .$
Work Step by Step
Since $ y=\sec^{-1}x $, then
$$ y'= \frac{1}{|x|\sqrt{x^2-1}}.$$
Now, we have
$$ y'(4)= \frac{1}{4\sqrt{16-1}}=\frac{1}{4\sqrt{ 15}} .$$
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