#d/(dx)(sec2x)=2sec(2x)tan(2x)#
#d/(dx)(tan2x)=2sec^2(2x)#
Short Answer
Use the derivatives of trig functions and the chain rule:
#d/(dx)(sec(2x))=sec(2x)tan(2x)*2=2sec(2x)tan(2x)#.
#d/(dx)(tan(2x))=sec^2(2x)*2=2sec^2(2x)#.
Explanation
You'll need the derivatives of #y=secx# and #y=tanx#
#d/(dx)(secx)=secxtanx#. and #d/(dx)(tanx)=sec^2x#
(Here's more on #d/(dx)(secx)# and #d/(dx)tanx# ).
You'll also want the Chain Rule.
There are various notations for derivatives and the Chain Rule, but for this question, this is a good one:
Suppose that we know #d/(dx)(f(x))=f'(x)#, then if we want #d/(dx)(f(u))# , the Chain Rule tells us to find the derivative of the outside function (that's #f'#) and evaluate it, not at #x#, but at #u#. Then multiply by the derivative of the inside.
The Chain Rule:
#d/(dx)(f(u))=f'(u)*(du)/(dx)#
Finding #d/(dx)(tan(2x))#
The outside function is #tan# and the inside function (the #u#) is #2x#.
The derivative of the tangent function is the square of the secant function.
#d/(dx)(f(u))=d/(dx)(tan(u))=sec^2(u)*(du)/(dx)#
As you get more experience, you'll simply write:
#d/(dx)(tan(2x))=sec^2(2x)*2=2sec^2(2x)#.
Finding #d/(dx)(sec(2x))#
The outside function is #sec# and the inside function (the #u#) is #2x#.
The derivative of #f(x)=secx# is the function (singular), #f'(x)=secxtanx#.
So the derivative of #f(2x)# is #f'(2x)*d/(dx)(2x)#, we write:
#d/(dx)(sec(2x))=sec(2x)tan(2x)*2=2sec(2x)tan(2x)#.
