Answer
$$ y'= -\frac{1}{2\sqrt{x}} sech(\sqrt{x}) \tanh (\sqrt{x}) .$$
Work Step by Step
Since $ y=sech(\sqrt{x})$, then the derivative, by using the chain rule and the facts that $(\ sech \ x)'= -\ sech \ x \tanh x $ and $(\sqrt{x})'=\frac{1}{2\sqrt{x}}$, is given by
$$ y'=-sech(\sqrt{x}) \tanh (\sqrt{x}) (\sqrt{x})'=- \frac{1}{2\sqrt{x}} sech(\sqrt{x}) \tanh (\sqrt{x}) .$$
