what is derivative of secx^2

$$\frac{d}{dx}sec^2x\\\\ =\frac{d}{dx}(cosx)^{-2}\\\\ =-2(cosx)^{-3}(-sinx)\\\\ =-2(cosx)^{-3}(-sinx)\\\\ =2(cosx)^{-3}(sinx)\\\\ =\frac{2sinx}{(cosx)^{3}}\\\\ or\\\\ =2tanx\;sec^2x$$

 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 !!!!!!

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!

You are a surgeon with numbers; just sharpen your machete :)

Me or Chris?

My "machete" made a hash of that one !!!!

Your "machete" is sharp enough, Melody.......mine needs to go back for a little more honing.......

Yea - I don't have a machete 

My tools are much more refined   LOL

You are not the first surgeon to k**l a patient. Remember they get to bury their mistakes;  mathematicians get to erace theirs. This is the lament: remembering to use a pencel and not a pen.