<s>[INST] Find the Derivative: This problem for my Calculus 1 class has got me stumped. I am not sure on where to start for this problem. Any help would be much appreciated. 
$y=x\tanh^{-1}(x) + \ln(\sqrt{1-x^2})$
 [/INST] $y=x\tanh^{−1}(x)+\ln(\sqrt{1−x^2})$
differentiation is linear
$\frac {dy}{dx} = \frac d{dx} (x\tanh^{−1}(x)) +\frac d{dx}(\ln(\sqrt{1−x^2})$
next we need to know the product rule and the chain rule
$\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac d{dx}\sqrt{1−x^2}\\
\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac {-x}{\sqrt{1−x^2}}\\
\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) -\frac {x}{1−x^2}$
Which leave one nasty left to tackle
$u = \tanh^{-1} x\\
\tanh u = x\\
\sech^2 u \frac {du}{dx} = 1\\
\frac {du}{dx} = \frac 1{\sech^2 u}\\
\frac {du}{dx} = \frac 1{1-\tanh^2 u}\\
\frac {du}{dx} = \frac 1{1- x^2}\\
$
$\frac {dy}{dx} = \tanh^{−1}(x) + \frac {x}{1-x^2} -\frac {x}{1−x^2}\\
\frac {dy}{dx} = \tanh^{−1}(x)$
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