# How do you differentiate f(x)= (3x+1)^4(4-x^2)  using the product rule?

Mar 9, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[- 2 x {\left(3 x + 1\right)}^{4}\right] + \left[\left(48 - 12 {x}^{2}\right) {\left(3 x + 1\right)}^{3}\right]$

#### Explanation:

Given -

$f \left(x\right) = {\left(3 x + 1\right)}^{4} \left(4 - {x}^{2}\right)$
$y = {\left(3 x + 1\right)}^{4} \left(4 - {x}^{2}\right)$

Use chain rule and product rule to differentiate this -

dy/dx=[(3x+1)^4(-2x)]+[(4-x^2)(4)(3x+1)^3(3)
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[- 2 x {\left(3 x + 1\right)}^{4}\right] + \left[12 \left(4 - {x}^{2}\right) {\left(3 x + 1\right)}^{3}\right]$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left[- 2 x {\left(3 x + 1\right)}^{4}\right] + \left[\left(48 - 12 {x}^{2}\right) {\left(3 x + 1\right)}^{3}\right]$