How do you differentiate $y = {(x + tan 2 π x)}^{2}$ $y = (x + tan 2pi x )^2$ ?

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Answer 1 · 2018-07-06T18:36:35

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

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

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

How do you differentiate y = (x + tan 2pi x )^2y=(x+tan2πx)2y = (x + tan 2pi x )^2?