# Differentiate in the form of wrtx? sin^3xsin3x

$\left({\sin}^{3} x \setminus \cdot \setminus \sin \left(3 x\right)\right) ' = 3 \setminus {\sin}^{2} x \left(\setminus \cos x \setminus \sin \left(3 x\right) + \setminus \sin x \setminus \cos \left(3 x\right)\right)$
We need to use the Product Rule , and the Chain Rule to differentiate $\setminus {\sin}^{3} x = {\left(\setminus \sin x\right)}^{3}$ and $\setminus \sin \left(3 x\right)$.
$\left({\sin}^{3} x \setminus \cdot \setminus \sin \left(3 x\right)\right) ' = \left(\setminus {\sin}^{3} x\right) ' \setminus \cdot \setminus \sin \left(3 x\right) + \left(\setminus {\sin}^{3} x\right) \setminus \cdot \left(\setminus \sin \left(3 x\right)\right) ' = 3 \setminus {\sin}^{2} x \setminus \cos x \setminus \sin \left(3 x\right) + \setminus {\sin}^{3} x \setminus \cdot \left(3 \setminus \cos \left(3 x\right)\right) = 3 \setminus {\sin}^{2} x \left(\setminus \cos x \setminus \sin \left(3 x\right) + \setminus \sin x \setminus \cos \left(3 x\right)\right)$