Question

how to derivate sin squared x by first principle

Answer 1

you can use
sin(x+h) = sin(x) cos(h) + cos(h) sin(x)

also
\[\lim_{x \rightarrow 0}\frac{ \sin(x) }{ x }=0\]
see http://www.khanacademy.org/math/differential-calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x

Answer 2

you also need to use the properties of taking the sum or product of limits
see
http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx#Extras_Limit_LimitProp

we begin with
\[ \frac{df}{dx}= \lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h } \\
= \lim_{h \rightarrow 0}\frac{ \sin^2(x+h)-\sin^2(x) }{ h } \\
= \lim_{h \rightarrow 0}\frac{ (\sin(x)\cos(h)+\sin(h)\cos(x))^2-\sin^2(x) }{ h } \\
= \lim_{h \rightarrow 0}\frac{ \sin^2(x)\cos^2(h)+\sin^2(h)\cos^2(x)+2 \sin(x)\cos(x)\sin(h)\cos(h)-\sin^2(x) }{ h } \\
=\lim_{h \rightarrow 0}\frac{ \sin^2(x)(\cos^2(h)-1)+\sin^2(h)\cos^2(x)+2 \sin(x)\cos(x)\sin(h)\cos(h) }{ h } \\
\]

now the idea is use the limit of x->0 , sin(x)/ x =1