Question

Find dy/dx for y=sec(1+x^2).

Answer 1

Solution:

http://screencast.com/t/NEYiosPUi

Answer 2

is it not the same 2 layers derivative ? (sec(u))' = u' sec u tan u = 2x sec (1+x^2) tan (1+x^2)?
either way is right, right?

Answer 3

Abselutely @Loser66 . I made a mistake . The derivetive of cos(1+x^2) should be

-2x sin(1+x^2) in my solution.

Answer 4

\[y=\sec \left( 1+x ^{2} \right)\]
differentiate w.r.t. x,
\[\frac{ dy }{ dx }=\sec \left( 1+x ^{2} \right)\tan \left(1 +x ^{2} \right)2x\]