How to differentiate y=sec 2x ?

##dy/dx=2sec2xtan2x##

Ideas to use: -> ##dy/dx=dy/(du)*(du)/dx## (this is the chain rule) -> ##d/dx(secx)=secxtanx## (Click here for an explanation of this result https://socratic.org/calculus/differentiating-trigonometric-functions/derivatives-of-y-sec-x-y-cot-x-y-csc-x) ->There will also be a substitution

How to do it: ##y=secu## where ##u=2x## ##dy/(du)=secutanu## and ##(du)/dx=2##

Recall ##dy/dx=dy/(du)*(du)/dx## Therefore ##dy/dx=secutanu*2=2sec2xtan2x##

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