What is the derivative of h(x)=sin⁡2xcos⁡2x

Step 1
Product Rule of differentiation:
If f(x) and g(x) are any two differentiable functions, then
${\displaystyle {f\left(g\right)}^{\prime }\left(x\right)=f\left(x\right)\times g^{\prime }\left(x\right)+f^{\prime }\left(x\right)\times g\left(x\right)}$
Chain rule of differentiation:
${\displaystyle \frac{d}{dx}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\times \frac{d}{dx}g\left(x\right)}$
${\displaystyle =f^{\prime }\left(g\left(x\right)\right)\times g^{\prime }\left(x\right)\times \frac{d}{dx}\left(x\right)}$
${\displaystyle =f^{\prime }\left(g\left(x\right)\right)\times g^{\prime }\left(x\right)}$
The given function is ${\displaystyle h\left(x\right)=\sin 2x\cos 2x}$
Step 2
Differentiate the given function using the product rule as follows.
${\displaystyle h^{\prime }\left(x\right)=\frac{d}{dx}\left(\sin 2x\cos 2x\right)}$
${\displaystyle =\sin 2x\frac{d}{dx}\left(\cos 2x\right)+\cos 2x\frac{d}{dx}\left(\sin 2x\right)}$
(U sin g product rule)
${\displaystyle =\sin 2x(-\sin 2x)\left(2\right)+\cos 2x\left(\cos 2x\right)\left(2\right)}$
(U sin g chain rule)
${\displaystyle =-2{\sin }^{2}2x+2{\cos }^{2}2x}$
${\displaystyle =-2(1-{\cos }^{2}2x)+2{\cos }^{2}2x}$
${\displaystyle =-2+4{\cos }^{2}2x}$
Therefore, the derivative of the given function is
${\displaystyle h^{\prime }\left(x\right)=-2+4{\cos }^{2}2x}$

Step 1
Given function:
${\displaystyle \sin \left(2x\right)\cos \left(2x\right)}$
${\displaystyle \frac{1}{2}\left(2\sin \left(2x\right)\cos \left(2x\right)\right)}$
${\displaystyle \frac{1}{2}\sin \left(4x\right)}$
Differentiating given function w.r.t x as follows
${\displaystyle \frac{d}{dx}\left(\frac{1}{2}\sin \left(4x\right)\right)}$
${\displaystyle =\frac{1}{2}\frac{d}{dx}\left(\sin \left(4x\right)\right)}$
${\displaystyle =\frac{1}{2}\cos \left(4x\right)\frac{d}{dx}\left(4x\right)}$
${\displaystyle =\frac{1}{2}\cos \left(4x\right)\left(4\right)}$
${\displaystyle =2\cos \left(4x\right)}$

Expalnation:
differentiate using the $\frac{\text{product}}{\text{chain rule}}$
$\begin{array}{}\text{Given}\\ y=f(x)g(x)\\ \frac{dy}{dx}=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)\Rightarrow \text{product rule}\\ f(x)=\sin 2x\Rightarrow f^{\prime }(x)=2\cos 2x\\ g(x)=\cos 2x\Rightarrow g^{\prime }(x)=-2\sin 2x\\ \frac{d}{dx}(\sin 2x\cos 2x)\\ =-2{\sin }^{2}2x+2{\cos }^{2}2x\\ =2({\cos }^{2}2x-{\sin }^{2}2x)=2\cos 4x\end{array}$