Using the definition of derivative to evaluate $(\sin^2(x))'$ I know that $\lim_{x\to 0} \dfrac{\sin x}{x} = 1$. But I'm stuck in using the definition of derivative to evaluate $(\sin^2 x)'$. Will appreciate any helpful input. Thank you.
 A: This is another way, which is quite similar but i think its worth showing.
$$(\textrm{sin}^2(x^2))'=\lim_{t\rightarrow x}\frac{f(x)-f(t)}{x-t}=\lim_{t\rightarrow x}\frac{\sin^2(x)-\sin^2(t)}{x-t}=$$
$$\lim_{t\rightarrow x}\frac{\sin(x+t)\sin(x-t)}{x-t}=\lim_{t\rightarrow x}\ \sin(x+t)\cdot\lim_{t\rightarrow x}\frac{\sin(x-t)}{x-t}=$$
$$\lim_{t\rightarrow x}\ \sin(x+t)\cdot\lim_{u\rightarrow 0}\frac{\sin(u)}{u}=
\lim_{t\rightarrow x}\ \sin(x+x)\cdot 1= \sin(2x)$$
A: $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
Hence,
\begin{align}
(\sin^2  x)' 
  &= \lim_{h\to 0}\frac{\sin^2(x+h)-\sin^2 (x)}{h} \\
  &=\lim_{h\to 0}\frac{\sin(2x+h)\sin(h)}{h} \\ 
  &=\lim_{h\to 0}\sin(2x+h) \cdot \lim_{h\to 0}\frac{\sin h}{h}=(\sin 2x) \cdot (1)=\sin2x.
\end{align}
In the second step,I have used the fact $(\sin^2A-\sin^2B)=\sin(A+B).\sin(A-B)$.
A: Implement the formula: 
1) $[f^n(x)]'=nf^{n-1}(x)\cdot f'(x)$ 
2) $(\sin(x))'=\cos(x)$
3) $\sin(x)\cos(x)=\sin(2x)$
we have:
$(\sin^2 (x))=2\sin(x)(\sin(x))'=2\sin(x)\cos(x)=\sin(2x)$
