#2**+10 **

$$\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

Melody
Oct 9, 2014

#1**+5 **

d/dx(secx)^{2} = (using the chain rule)

2(secx)(tanx)^{2}

----------------------------------------------------------------------------------------------------------------

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 !!!!!!

CPhill
Oct 9, 2014

#2**+10 **

Best Answer

$$\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

Melody
Oct 9, 2014

#3**0 **

It wasn't much of an error Chris, you didn't need to take your points away! I'll give them back to you. :)

Melody
Oct 9, 2014

#4**0 **

Nope....I'm "zapping" myself for that stupid error......I should have checked my work more closely !!!!

CPhill
Oct 9, 2014

#5**0 **

You can't take away my points to you. It is not polite to throw a gift in the bin!

Melody
Oct 9, 2014