How do you differentiate $sin^2(x/6)$ ?

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Answer 1 · 2015-08-19T09:54:58

$y^' = 1/3 * sin(x/6) * cos(x/6)$

You can differentiate this function

$y = sin(x/6)$

by using the chain rule twice, once for $u_1^2$ , with $u_1 = sin(x/6)$ , and once for $sinu_2$ , with $u_2 = x/6$ .

Your target derivative will be

$d/dx(y) = d/(du_1)(u_1^2) * d/dx(u_1)$ , with

$d/dxu_1 = d/(du_2) * sinu_2 * d/dx(u_2)$

This will give you

$d/dx(u_1) = cosu_2 * d/dx(x/6)$

$d/dx(sin(x/6)) = cos(x/6) * 1/6$

Plug this back into your target derivative to get

$y^' = 2u_1 * 1/6 * cos(x/6)$

$y^' = color(green)(1/3 * sin(x/6) * cos(x/6))$

How do you differentiate sin^2(x/6)sin^2(x/6)?