Question #ec712

1 Answer
Apr 7, 2017

ddxtan2x=2tanxsec2x

ddxsec2x=2sec2xtanx

Explanation:

ddxtan2x

Apply chain rule,

r(u)r1ddu

=2tanxddx(tanx)

ddxtanx=sec2x (common derivative)

Therefore the final answer is,

ddxtan2x=2tanxsec2x


ddxsec2x

Apply chain rule

r(u)r1ddu

2secxddxsecx

ddxsecx=secxtanx (common derivative)

Therefore the final answer is,

ddxsec2x=2secxsecxtanx

Which simplifies to,

ddxsec2x=2sec2xtanx