# How do you differentiate sin²(2x)?

Jul 17, 2015

This is nothing crazy complicated if you get the pattern; you can do the following (I presented this in regular prime notation and function composition notation):

$\textcolor{g r e e n}{\text{Overall Function}}$
$\textcolor{p u r p \le}{\text{First Inner Function}}$
$\textcolor{\mathrm{da} r k b l u e}{\text{Innermost Function}}$
$\textcolor{h i g h l i g h t}{\text{Derivative}}$

${\sin}^{2} \left(2 x\right) = {\left(\sin \left(2 x\right)\right)}^{2}$

Let:
$\textcolor{g r e e n}{f \left\{g \left[h \left(x\right)\right]\right\}} = f \left[\left(g \circ h\right) \left(x\right)\right] \textcolor{g r e e n}{= {\left\{\textcolor{p u r p \le}{g \left[h \left(x\right)\right]}\right\}}^{2}}$
$\textcolor{p u r p \le}{g \left[h \left(x\right)\right]} = \left(g \circ h\right) \left(x\right) \textcolor{p u r p \le}{= \sin \left[\textcolor{\mathrm{da} r k b l u e}{h \left(x\right)}\right]}$
$\textcolor{\mathrm{da} r k b l u e}{h \left(x\right) = 2 x}$

$\frac{d}{\mathrm{dx}} \left(\textcolor{g r e e n}{f \left\{g \left[h \left(x\right)\right]\right\}}\right) = \textcolor{h i g h l i g h t}{f '} \left\{\textcolor{p u r p \le}{g \left[h \left(x\right)\right]}\right\} \cdot \textcolor{h i g h l i g h t}{g '} \left[\textcolor{\mathrm{da} r k b l u e}{h \left(x\right)}\right] \cdot \textcolor{h i g h l i g h t}{h ' \textcolor{b l a c k}{\left(x\right)}}$

$= \textcolor{h i g h l i g h t}{2} \left[\textcolor{p u r p \le}{\sin \left(2 x\right)}\right] \cdot \textcolor{h i g h l i g h t}{\cos} \left(\textcolor{\mathrm{da} r k b l u e}{2 x}\right) \cdot \textcolor{h i g h l i g h t}{2}$

$= 2 \left(2 \sin \left(2 x\right) \cos \left(2 x\right)\right)$

$= \textcolor{b l u e}{2 \sin \left(4 x\right)}$

...since $\sin \left(2 x\right) = 2 \sin x \cos x$ and thus $\sin \left(4 x\right) = 2 \sin \left(2 x\right) \cos \left(2 x\right)$.