Answer
$y'=\tanh^{-1}x$
Work Step by Step
Given: $y=x\tanh^{-1}x+\ln{\sqrt{1-x^2}}$
Now, $y'=\dfrac{d}{dx}(x\tanh^{-1}x+\ln{\sqrt{1-x^2}})$
This implies $y'=\dfrac{d}{dx}x\tanh^{-1}x+\dfrac{d}{dx}\ln\sqrt{1-x^2}$
Apply product rule, we have
$y'=(\tanh^{-1}x+x\dfrac{d}{dx}\tanh^{-1}x)+\dfrac{d}{dx}\ln{\sqrt{1-x^2}}$
$=(\tanh^{-1}x+x)(\dfrac{1}{1-x^2}+\dfrac{d}{dx}\ln{\sqrt{1-x^2}})$
or, $=\tanh^{-1}x+\dfrac{x}{1-x^2}+\dfrac{d}{dx}\ln{\sqrt{1-x^2}}$
Now, apply chain rule.
$y'=\tanh^{-1}x+\dfrac{x}{1-x^2}+\dfrac{d\ln{\sqrt{1-x^2}}}{d({\sqrt{1-x^2})}}(\dfrac{d({\sqrt{1-x^2})}}{d1-x^2})(\dfrac{d(1-x^2)}{dx})$
or, $=\tanh^{-1}x+\dfrac{x}{1-x^2}+\dfrac{1}{\sqrt{1-x^2}}(\dfrac{1}{2\sqrt{1-x^2}})( -2x)$
or, $=\tanh^{-1}x+\dfrac{x}{1-x^2}+\dfrac{-x}{{1-x^2}}$
Thus, $y'=\tanh^{-1}x$
