What is the derivative of #sec^2 2x#?

Question

19601 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-05T15:54:01

#4sec^2 2xtan2x#

Use the chain rule. The first issue is the squared term.

#d/dx(u^2)=2u*u'#

#d/dx(sec^2 2x)=2sec2x*d/dx(sec2x)#

Use chain rule again:

#d/dx(secu)=secutanu*u'#

#d/dx(sec^2 2x)=2sec2x*sec2xtan2x*d/dx(2x)#

Since #d/dx(2x)=2#,

#d/dx(sec^2 2x)=4sec^2 2xtan2x#