Answer
$- sech^{-1}x \dfrac{x}{\sqrt{1- x^2}}$
Work Step by Step
We are given: $y=\ln x+\sqrt {1-x^2} sech^{-1}x$
Recall the formula: $\dfrac{d (sech^{-1} x)}{dx}=\dfrac{-1}{x \sqrt{1- x^2}}$
We need to use the product rule to get the differentiation:
Thus, $\dfrac{dy}{dx}=\dfrac{1}{x}+[(\sqrt {1-x^2}) (\dfrac{-1}{x \sqrt{1- x^2}}) + sech^{-1}x (\dfrac{1}{ 2\sqrt{1- x^2}})(-2x)] =\dfrac{1}{x} - \dfrac{\sqrt {1-x^2}}{x \sqrt{1- x^2}} - sech^{-1}x \dfrac{x}{\sqrt{1- x^2}}=- sech^{-1}x \dfrac{x}{\sqrt{1- x^2}}$$