What is the derivative of #(sin(pix))^2#?

1 Answer
Mar 23, 2017

#2 pi sin pi x cos pi x = pi sin 2pix#

Explanation:

Use the power rule #(u^n)' = n u^(n-1)*u'# and
the chain rule #(sin v)' = v' cos v#

Let #u = sin (pi x)#, #v = pi x#, #v' = pi#

So #u' = (sin pi x)' = pi cos pi x#

#((sin pi x)^2)' = 2(sin (pix)) (pi cos pi x) = 2 pi sin pi x cos pi x#

Use the Double Angle Formula #sin 2u = 2 sin u cos u#:

#2 pi sin pi x cos pi x = pi sin 2pix#