How to evaluate $\frac{d}{dx}\sin^2x$ I'm having trouble understanding how  $\frac{d}{dx}\sin^2x =2\sin(x)\cos(x)$.
Please show as many steps of the proof as necessary so that I can apply this to other problems.
Thank you for your time~! ^_^
 A: $\frac{d}{dx}\sin^2x= 2\sin(x)(\sin(x))'= 2\sin(x)\cos(x)$
A: We have that by the product rule,
$$\begin{align}\frac{d}{dx}\sin^2(x)&=\frac{d}{dx}(\sin(x)\sin(x))\\&=
(\frac{d}{dx}\sin(x))\sin(x)+\sin(x)(\frac{d}{dx}\sin(x))\\&=
\cos(x)\sin(x)+\sin(x)\cos(x)\\&=2\sin(x)\cos(x).\end{align}$$
A: $$\dfrac{d(\sin^nx)}{dx}=\dfrac{d(\sin^nx)}{d(\sin x)}\cdot\dfrac{d(\sin x)}{dx}=?$$
See Chain rule
A: You have a composite function
$$
x\quad\color{red}{\longrightarrow}\quad u=\sin x\quad\color{blue}{\longrightarrow}\quad y=u^2.
$$


*

*The derivative of $u\color{blue}\to u^2$ is $2u$, but $u=\sin x$, then it is $\color{blue}{2\sin x}$.

*The derivative of $x\color{red}\to\sin x$ is $\color{red}{\cos x}$.


The chain rule says: take their product
$$
\color{blue}{2\sin x}\cdot\color{red}{\cos x}.
$$
A: With bare hands:
$$\sin^2x=\frac{1-\cos2x}2$$ so that
$$(\sin^2x)'=-\frac12(-2\sin2x)=\sin2x=2\sin x\cos x.$$ 
A: The derivative of a function $(f(x))^n$ is given by 
$$ ((f(x))^n)' = \frac{d (f(x))^n }{ d f(x)} \frac{d f(x)}{dx} = n f(x)^{n-1} f'(x), $$
by the chain rule. 
Hence for your example $(f(x))^2 = (\sin(x))^2$ and the derivative is 
$$\left((\sin(x))^2 \right)' = 2 \sin(x) \cdot (\sin(x))' = 2 \sin(x) \cos(x). $$
A: Using the definition of the derivative and the general theorem $\lim(fg)=(\lim f)(\lim g)$ provided $\lim f$ and $\lim g$ both exist, we have
$$\begin{align}
{d\over dx}\sin^2x
&=\lim_{y\to x}{\sin^2x-\sin^2y\over x-y}\\
&=\left(\lim_{y\to x}(\sin x+\sin y)\right)\left(\lim_{y\to x}{\sin x-\sin y\over x-y} \right)\\
&=(\sin x+\sin x)\left({d\over dx}\sin x\right)\\
&=2\sin x\cos x
\end{align}$$
A: $$\frac{{\rm d}(\sin^2 x)}{{\rm d}x}=\frac{{\rm d}(\sin^2 x)}{{\rm d}(\sin x)}\cdot \frac{{\rm d}(\sin x)}{{\rm d}x}=2\sin x \cos x.$$
