So I'm taking derivatives of trig functions.  I know that if y=sin(x), then y'=cos(x).  I'm wondering if you know calculus, what is y' if y=sin²(x)?  I'm not sure if I'm supposed to use the power rule(dy/dx xⁿ = nx^n-1) and/or the chain rule(dy/dx f(g(x)) = f'(g(x))g'(x)).

Assuming you mean sin(2x)... {{not sin^2(x)...}} it'd be 2cos(2x).

Let's work through it.

You can always substitute one variable for another if you're in a bind... so let's sub u for 2x.

So -- what's the derivative of sin(u)... well, that's be cos(u) as you just said.

Okay -- but the chainrule says we need to multiply by the derivative of u.

Well -- since u = 2x... what's the derivative of 2x?

Well... that's 2.
... So we multiply cos(u) by 2.
... which is 2cos(u)
... which is 2cos(2x)

... if you can do it by eye (which -- yea you eventually definitely will be able to) then you don't need to do substitutions... but that's just easier to read. (and is something you're going to be doing FOREVER)

If you actually did mean sine_squared(x) -- it would be 2sin(x)cos(x).

The power rule and the chainrule are actually the same thing, sorta...

{sin(x)}^2... sub sin(x) for "u".

u^2 differentiates to 2u. Well -- you need to multiply by the derivative of u again, by the chainrule... what's the derivative of u?... well what's the derivative of sin(x)? ... cos(x).

Therefore -- 2*u*cos(x) which is 2*sin(x)*cos(x).

If you wish me to explain it better -- I'll do it tomorrow after class.

Obviously "^" is the sign for "to the power of"...

wheres my gun..

help, everything was absolutely clear to me! 

Something's wrong with me, isn't it? ;_;