How do you derive a function where the trig is under a square root(adsbygoogle = window.adsbygoogle || []).push({});

like this:

y=sqr(tan^-1(x))

y=(tan^-1(x))^(1/2)

i know that y=tan^-1u(x) => dy/dx=1/(1+u^2)*du/dx

But how do I work it out? The book doesn't give the answers to pair numbers :(

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# Homework Help: D/dx of Inverse trig

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