https://ocw.mit.edu/courses/mathema...nd-inverse-functions/MIT18_01SCF10_Ses15b.pdf(adsbygoogle = window.adsbygoogle || []).push({});

so derivative of arctan is 1/(x^2+1) and this is obvious since we are considering trigonometry of a triangle of which hypotenuse is sqrt(x^2+1).

But can we consider its hypotenuse to be x instead of sqrt(x^2+1)? where cosy=1/x

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# B Really an easy question about derivative of arctan

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