Answer
$-\displaystyle \frac{x\cdot{\rm sech}^{-1}x}{\sqrt{1-x^{2}}}$
Work Step by Step
$y=\ln x+(1-x^{2})^{1/2}{\rm sech}^{-1}x$
$\displaystyle \frac{d}{dx}[\ln x]=\frac{1}{x}$
$\displaystyle \frac{d}{dx}[(1-x^{2})^{1/2}]$
Apply the chain rule:
$=\displaystyle \frac{1}{2}\cdot(1-x^{2})^{-1/2}\cdot(-2x)=\frac{-x}{\sqrt{1-x^{2}}}$
$\displaystyle \frac{d}{dx}[{\rm sech}^{-1}x]=\frac{-1}{x\sqrt{1-x^{2}}}\quad$ ... (table 10)
Applying the sum and product rules of differentiation
$\displaystyle \frac{dy}{dx}=\\\frac{d}{dx}[\ln x]+(1-x^{2})^{1/2}\cdot\frac{d}{dx}[{\rm sech}^{-1}x]+\frac{d}{dx}[(1-x^{2})^{1/2}]\cdot{\rm sech}^{-1}x$
$=\displaystyle \frac{1}{x}+(1-x^{2})^{1/2}(\frac{-1}{x\sqrt{1-x^{2}}})-\frac{x}{\sqrt{1-x^{2}}}\cdot{\rm sech}^{-1}x$
$=\displaystyle \frac{1}{x}-\frac{1}{x}-\frac{x{\rm sech}^{-1}x}{\sqrt{1-x^{2}}}$
= $-\displaystyle \frac{x{\rm sech}^{-1}x}{\sqrt{1-x^{2}}}$
