# Differentiate tan−1   [ (1+x2)1/2 - (1-x2)1/2 ]  w.r.t x ?                               (1+x2)1/2 + (1-x2)1/2

Pankaj
8 years ago
I am not able to interpret your query. Please write down the problem clearly so that I can provide you the meaningful answer.
Thanks & Regards
Pankaj Singh
$\\\frac{d}{du}\left(\arctan \left(u\right)\right) \\\frac{1}{u^2+1}x\left(\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{1-x^2}}\right) \\u=\sqrt{1+x^2}-\sqrt{1-x^2} \\=\frac{1}{\left(\sqrt{1+x^2}-\sqrt{1-x^2}\right)^2+1}x\left(\frac{1}{\sqrt{x^2+1}}+\frac{1}{\sqrt{1-x^2}}\right) \\=\frac{x\left(\sqrt{1-x^2}+\sqrt{x^2+1}\right)}{2x^4+3\sqrt{1-x^4}-2}$