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

Jun 26, 2018

$2 {x}^{3} \cdot 2 \cdot \cos {x}^{2} \cdot - \sin {\left(x\right)}^{2} \cdot 2 x + {\cos}^{2} {x}^{2} \cdot 6 {x}^{2}$

Explanation:

we have
$f \left(x\right) = \left(2 {x}^{3}\right) \left({\cos}^{2} {x}^{2}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{3} \cdot d \frac{{\cos}^{2} {x}^{2}}{\mathrm{dx}} + \left({\cos}^{2} {x}^{2}\right) \cdot d \frac{2 {x}^{3}}{\mathrm{dx}}$

$\implies 2 {x}^{3} \cdot 2 \cdot d \frac{\cos {x}^{2}}{\mathrm{dx}} + \left({\cos}^{2} {x}^{2}\right) \cdot 2 \cdot 3 \cdot {x}^{2}$.....using$\left[\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{n} \implies n {x}^{n - 1}\right]$
$\implies 2 {x}^{3} \cdot 2 \cdot \left(\cos {x}^{2}\right) \cdot - \sin d \frac{{x}^{2}}{\mathrm{dx}} + \left({\cos}^{2} {x}^{2}\right) \cdot 6 \cdot {x}^{2}$
.........using $\left[\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \implies - \sin x\right]$
$\implies 2 {x}^{3} \cdot 2 \cdot \left(\cos {x}^{2}\right) \cdot - \sin {x}^{2} \cdot 2 x + \left({\cos}^{2} {x}^{2}\right) \cdot 6 \cdot {x}^{2}$