# How do you use the Product Rule to find the derivative of f(z) = z^3sin^2(2z)?

Oct 19, 2015

See the explanation.

#### Explanation:

I assume that he challenge is to differentiate ${\sin}^{2} \left(2 z\right)$

Note that ${\sin}^{2} \left(2 z\right) = {\left[\sin \left(2 z\right)\right]}^{2}$

So the derivative is
$2 \left[\sin \left(2 z\right)\right] \frac{d}{\mathrm{dz}} \left(\sin \left(2 z\right)\right) = 2 \left[\sin \left(2 z\right)\right] \left(\cos \left(2 z\right) \frac{d}{\mathrm{dz}} \left(2 z\right)\right)$

 = 2[sin(2z)] (cos(2z) (2)

$= 4 \sin \left(2 z\right) \cos \left(2 z\right)$

This last expression can be rewritten as $2 \sin \left(4 x\right)$

Now to the big question:

$f ' \left(z\right) = 3 {z}^{2} {\sin}^{2} \left(2 z\right) + {z}^{3} \left[2 \sin \left(2 z\right) \cos \left(2 z\right) \cdot 2\right]$

$= 3 {z}^{2} {\sin}^{2} \left(2 z\right) + 4 {z}^{3} \sin \left(2 z\right) \cos \left(2 z\right)$

Rewrite/simplify to taste.