Question

solve attached please .
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Answer 1

You may use $\tanh^{-1}(x)=\frac{1}{2}\ln\frac{1+x}{1-x}$ to simplify the work slightly :

$y=\log\left(\sqrt{\frac{1+x}{1-x}}\right)^{1/2}-\frac{1}{2}\tan^{-1}x$

$=\log\left(\frac{1+x}{1-x}\right)^{1/4}-\frac{1}{2}\tan^{-1}x$

$=\frac{1}{4}\log\frac{1+x}{1-x}-\frac{1}{2}\tan^{-1}x$

$=\frac{1}{2}\tanh^{-1}x-\frac{1}{2}\tan^{-1}x$

Now take the derivative 
$\frac{dy}{dx}=\frac{1}{2}\frac{1}{1-x^2}-\frac{1}{2}\frac{1}{1+x^2}$

$=\frac{x^2}{1-x^4}$