What is the derivative of y=sec(2x) tan(2x)y=sec(2x)tan(2x)?

1 Answer
WADU HEK
Apr 14, 2018

2sec(2x)(sec^2(2x) + tan^2(2x))(sec2(2x)+tan2(2x))

Explanation:

y' = (sec(2x))(tan(2x))' + (tan(2x))(sec(2x))' (Product Rule)
y' = (sec(2x))(sec^2(2x))(2) + (tan(2x))(sec(2x)tan(2x))(2) (Chain rule and derivatives of trig)
y' = 2sec^3(2x) + 2sec(2x)tan^2(2x)
y' = 2sec(2x)(sec^2(2x) + tan^2(2x))

Related questions
See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x)
Impact of this question
8040 views around the world
You can reuse this answer
Creative Commons License