How do you differentiate $y=sqrt(tan^-1x)$ ?

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Answer 1 · 2015-04-16T16:17:28

$dy/dx=1/(2sqrt(tan^-1x))(1/(1+x^2))$

$y=sqrt(tan^-1x)$

$dy/dx=1/2(tan^-1x)^((-1)/2)d/dxtan^-1x$

$dy/dx=1/(2sqrt(tan^-1x))(1/(1+x^2))$

How do you differentiate y=sqrt(tan^-1x)y=sqrt(tan^-1x)?