$$\frac{d}{dx}sec^2x\\\\
=\frac{d}{dx}(cosx)^{-2}\\\\
=-2(cosx)^{-3}(-sinx)\\\\
=-2(cosx)^{-3}(-sinx)\\\\
=2(cosx)^{-3}(sinx)\\\\
=\frac{2sinx}{(cosx)^{3}}\\\\
or\\\\
=2tanx\;sec^2x$$
or
$$\\2tanx(tan^2x+1)\\\\
or\\\\
2tan^3x+2tanx$$
I got a different answer from Chris - I haven't checked it
d/dx(secx)2 = (using the chain rule)
2(secx)(tanx)2
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Edit..........
Whoops...... I made a slight error......
The derivative of the sec x = sec xtanx...so the final result should be, as Melody said,
2(secx)(secxtanx) = 2tanx(secx)^2
Thanks, Melody, for catching my error !!!!!!
$$\frac{d}{dx}sec^2x\\\\
=\frac{d}{dx}(cosx)^{-2}\\\\
=-2(cosx)^{-3}(-sinx)\\\\
=-2(cosx)^{-3}(-sinx)\\\\
=2(cosx)^{-3}(sinx)\\\\
=\frac{2sinx}{(cosx)^{3}}\\\\
or\\\\
=2tanx\;sec^2x$$
or
$$\\2tanx(tan^2x+1)\\\\
or\\\\
2tan^3x+2tanx$$
I got a different answer from Chris - I haven't checked it
It wasn't much of an error Chris, you didn't need to take your points away! I'll give them back to you. :)
Nope....I'm "zapping" myself for that stupid error......I should have checked my work more closely !!!!
You can't take away my points to you. It is not polite to throw a gift in the bin!