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2015-05-04T17:19:04+05:30

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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}
2 5 2
we did tanh^-1(3x-2) in previous problem right ?
oh sorry i messed up i got it now !
it would be a good exercise to work it using chain rule and see both answers match
done.
dont mind, but r u a student or sir ?