How do you differentiate f(x) =xsec4x^3  using the chain rule?

Oct 5, 2017

$f ' \left(x\right) = 12 {x}^{3} \sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) + \sec \left(4 {x}^{3}\right)$

Explanation:

We seek $f ' \left(x\right)$ where:

$f \left(x\right) = x \sec \left(4 {x}^{3}\right)$

First we apply the product rule:

$f ' \left(x\right) = \left(x\right) \left(\frac{d}{\mathrm{dx}} \sec \left(4 {x}^{3}\right)\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left(\sec \left(4 {x}^{3}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = x d \left(\frac{d}{\mathrm{dx}} \sec \left(4 {x}^{3}\right)\right) + \left(1\right) \left(\sec \left(4 {x}^{3}\right)\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = x d \left(\frac{d}{\mathrm{dx}} \sec \left(4 {x}^{3}\right)\right) + \sec \left(4 {x}^{3}\right)$

Then, we apply the chain rule:

$f ' \left(x\right) = x \left(\sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) \frac{d}{\mathrm{dx}} \left(4 {x}^{3}\right)\right) + \sec \left(4 {x}^{3}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = x \sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) \left(12 {x}^{2}\right) + \sec \left(4 {x}^{3}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 12 {x}^{3} \sec \left(4 {x}^{3}\right) \tan \left(4 {x}^{3}\right) + \sec \left(4 {x}^{3}\right)$