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Home  >>  EAMCET  >>  Mathematics
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$y= \log \bigg \{ \bigg (\large\frac{1+x}{1-x}\bigg )^{1/4}\bigg\}-\frac{1}{2}$$ \tan ^{-1}(x),$ then $\large\frac{dy}{dx}$ is equal to

\[\begin {array} {1 1} (a)\;\frac{x}{1-x^2} & \quad (b)\;\frac{x^2}{1-x^4} \\ (c)\;\frac{x}{1+x^4} & \quad (d)\;\frac{x}{1-x^4} \end {array}\]

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1 Answer

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$(b)\;\frac{x^2}{1-x^4}$
answered Nov 7, 2013 by pady_1
 
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