Question

What is the derivative of `f(x)=1/3sec^3(2x)tan(2x)-2sec(2x)tan(2x)`?

Answer 1

`f'(x)=2sec^3(2x)tan(2x)-2sec(2x)tan(2x)`

Explanation:

Evaluate using the chain rule: `d/(dx)f(g(x))=f'(g(x))g'(x)`

`f(x)=1/3sec^3(2x)-sec(2x)`

`f'(x)=1/3(3)sec^2(2x)sec(2x)tan(2x)(2)-sec(2x)tan(2x)(2)`

Simplifying:

`=>f'(x)=2sec^3(2x)tan(2x)-2sec(2x)tan(2x)`