Question

find y'(1) if y=tan^-1x + sec^-1 square root of 1+x^2?

Answer 1

\[\tan ^{-1} x=1/(1+x ^{2})\]
\[\sec ^{-1}x=1/ x \sqrt{}(x ^{2}-1)\]

Answer 2

\[(\cos^{-1} (\sqrt{1 + x ^{2}})' = - x/ (\sqrt{1+x ^{2}} \sqrt{1-x ^{2})}\]\[[\sec ^{-1} (\sqrt{1 + x ^{2}})]' = 1/\cos^{-1} (\sqrt{1 + x ^{2}}) \]