How do you differentiate sin^3(x)cos(x)?

May 20, 2016

$3 {\sin}^{2} x . {\cos}^{2} x - {\sin}^{4} x$

Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus {\sin}^{3} \left(x\right) \setminus \cos \setminus \left(x\right)\right)$

Applying product rule,

${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = \setminus {\sin}^{3} \left(x\right) , \setminus g = \setminus \cos \left(x\right)$

$= \setminus \frac{d}{\mathrm{dx}} \left(\setminus {\sin}^{3} \left(x\right)\right) \setminus \cos \left(x\right) + \setminus \frac{d}{\mathrm{dx}} \setminus \left(\setminus \cos \left(x\right) \setminus r i g h t\right) \setminus {\sin}^{3} \left(x\right)$

We have,
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus {\sin}^{3} \setminus \left(x\right)\right) = 3 \setminus {\sin}^{2} \left(x\right) \setminus \cos \left(x\right)$
Also,
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \cos \left(x\right)\right) = - \setminus \sin \left(x\right)$

$= 3 \setminus {\sin}^{2} \left(x\right) \setminus \cos \left(x\right) \setminus \cos \left(x\right) + \left(- \setminus \sin \left(x\right)\right) \setminus {\sin}^{3} \left(x\right)$

Simplifying it,we get,
$= 3 \setminus {\sin}^{2} \left(x\right) \setminus {\cos}^{2} \left(x\right) - \setminus {\sin}^{4} \left(x\right)$