Answer
$$ f'(x)
=\frac{e^{\sec^{-1}x}}{|x|\sqrt{x^2-1}} .$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(\sec^{-1} x)'=\dfrac{1}{|x|\sqrt{x^2-1}}$
Since $ f(x)=e^{\sec^{-1}x}$, then the derivative, using the chain rule, is given by
$$ f'(x)=e^{\sec^{-1}x}(\sec^{-1}x)'=e^{\sec^{-1}x}\frac{1}{|x|\sqrt{x^2-1}} \\
=\frac{e^{\sec^{-1}x}}{|x|\sqrt{x^2-1}} .$$
