What is the rule behind this derivative? $$\dfrac{\rm d}{{\rm d}t}\big(\sin^2(t)\big)=\sin(2t).$$
I don't understand what is the rule behind this derivation.
I had tried to first rerivate sin() and then to derivate the square function, but apparently that's the wrong way.
 A: We use the chain rule (as you clearly seem to have done), and after doing so, we recall that $$2\sin(t)\cos(t) = \sin t \cos t + \sin t \cos t = \sin(t + t) = \sin (2t)$$ (The double angle formula for $\sin (2t)$ is derived from the angle sum identity, as you see.)
$$\begin{align}\dfrac{\rm d}{{\rm d}t}\big(\sin^2(t)\big)& = \frac d{dt}\Big((\sin(t))^2\Big)\\ \\ 
& = 2\sin(t) \cdot \frac d{dt}(\sin t) \\ \\
& = 2\sin(t)\cos(t) \\ \\
& = \sin(2t)\end{align}$$
A: Trig identities are always your friend
The easiest way to do this would be to recognize that $$\sin^2(t)=\frac{1-\cos(2x)}{2}$$And then try doing the derivative for that, which would also be much much simpler. (The constant cancels out and you get $\sin(2x)/2$, and you multiply the whole thing by $2$ to get $\sin(2x)$
A: You are right, $\frac{d}{dt}\sin^2t=2\sin t\cos t$.
By using the trig identity $\sin(2t)=2\sin t\cos t$, we can simplify this to $\sin(2t)$, so:
$$\frac{d}{dt}\sin^2t=2\sin t\cos t=\sin(2t)$$
A: From the defintion,
$$F=\frac{d(\sin^2t)}{dx}=\lim_{h\to0}\frac{\sin^2(t+h)-\sin^2t}{t+h-t}$$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$F=\lim_{h\to0}\frac{\sin(t+h-t)\sin(t+h+t)}h=\lim_{h\to0}\frac{\sin h}h\cdot\lim_{h\to0}\sin(2t+h)=?$$
