# How do you find the derivative of g(x)= 3tan4xsin2xcos2x?

May 14, 2015

First just remember

$\sin \left(a\right) \cdot \cos \left(a\right) = \frac{1}{2} \sin \left(2 a\right)$

$\frac{3}{2} \tan \left(4 x\right) \sin \left(4 x\right)$

$= \frac{3}{2} {\sin}^{2} \frac{4 x}{\cos} \left(4 x\right)$

Remember ${\sin}^{2} \left(a\right) = 1 - {\cos}^{2} \left(a\right)$

$\frac{3}{2 \cos \left(4 x\right)} - \frac{3}{2} \cos \left(4 x\right)$

$\left(\frac{3}{2} {\cos}^{-} 1 \left(4 x\right)\right) ' = \frac{3}{2} \left({\cos}^{-} 1 \left(4 x\right)\right) ' = 6 \sin \left(4 x\right) \cdot {\cos}^{-} 2 \left(4 x\right)$

$\left({u}^{n}\right) ' = n \cdot u ' \cdot {u}^{n - 1}$

$\left(\frac{3}{2} \cos \left(4 x\right)\right) ' = \frac{3}{2} \left(\cos \left(4 x\right)\right) ' = - 6 \sin \left(4 x\right)$

So the result is

$= 6 \sin \left(4 x\right) \cdot {\cos}^{-} 2 \left(4 x\right) + 6 \sin \left(4 x\right)$

$6 \sin \left(4 x\right) \left({\cos}^{-} 2 \left(4 x\right) + 1\right)$