Derivative of y=sin^2x: Need Help!

[hummer]

Hi there. Need a bit of help~

y=sin^(2)x

So the equation reads y equals sine squared times x

Anyway, my teacher told me that in cases like these, you can move the exponent "outside," so that the equation can be written as y=(sinx)^2.

So on to getting the derivative of the equation:

I did the package rule, so y'=2(sinx)cosx

Unfortunately, the answer on the worksheet I have is sin2x. No cosine. Am I not supoosed to do the package rule?? Or could it be simplified?? Or am I just doing something else wrong completely?

Help~ TT__TT

 

[danago]

Have a look at the common trig identities Shouldnt be hard to see how they got to the next step.

 

[Feldoh]

You're answer is right, but as danago said look at some trig identities, and it should seem very clear :P

 

[hummer]

Heh... after staring at some trig identities that I've never saw before for about 20 minutes, I finally got it through my thick head what you guys were talking about... X/ Oh dear... Thanks a bunch!

 

[HallsofIvy]

First of all $sin^2(x)$ does not read "y equals sine squared times x"! That's either very sloppy wording or you are completely misunderstanding. That notation means "First find sine of x. Then square hat." You are certainly not multiplying "sine" (which is a function not a value) by anything.