Question

How do you differentiate `y = (x + tan 2pi x )^2`?

Answer 1

`y'=2(x+tan(2pix))(1+2pi*(1+tan^2(2pix))`

Explanation:

We use that

`(tan(x))'=1/cos^2(x)=(sin^2(x)+cos^2(x))/cos^2(x)=1+tan^2(x)`
and
(x^n)'=nx^(n-1)#
and the chain rule:

`(f(g(x)))'=f'(g(x))*g'(x)`
so we get

`y'=2(x+tan(2pix))(1+2pi(1+tan^2(2pix)))`