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))#