How do you differentiate f(z) = z^3sin^2(2z) using the product rule?

Mar 17, 2016

Derivative of $f \left(z\right)$ is ${z}^{2} \sin \left(2 z\right) \left\{4 z \cos \left(2 z\right) + 3 \sin \left(2 z\right)\right\}$

Explanation:

Product rule states that derivative of a function $f \left(x\right)$ which is a product of two functions $g \left(x\right)$ and $h \left(x\right)$ i.e. $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$ is given by

$\frac{d}{\mathrm{dx}} f \left(x\right) = g \left(x\right) \cdot \frac{d}{\mathrm{dx}} h \left(x\right) + \frac{d}{\mathrm{dx}} g \left(x\right) \cdot h \left(x\right)$

We will also need here the concept of derivative of a function of a function, say $\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dg}} \cdot \frac{\mathrm{dg}}{\mathrm{dx}}$

Hence, derivative of $f \left(z\right) = {z}^{3} {\sin}^{2} \left(2 z\right)$ is given by

$\frac{d}{\mathrm{dx}} f \left(z\right) = {z}^{3} \cdot \frac{d}{\mathrm{dx}} {\sin}^{2} \left(2 z\right) + \frac{d}{\mathrm{dx}} {z}^{3} \cdot {\sin}^{2} \left(2 z\right)$

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

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

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